# 2.3. Sequences and Series#

This section collects results on sequences and series of real numbers.

Definition 2.28

A sequence of real numbers is a function $$f : \Nat \to \RR$$.

A sequence can be thought of as an ordered (countable) list of real numbers.

Definition 2.29 (Convergence)

A sequence $$\{ x_n \}$$ of real numbers is said to converge to $$x \in \RR$$ if for every $$\epsilon > 0$$, there exists a natural number $$n_0$$ (depending upon $$\epsilon$$) such that

$| x_n - x | < \epsilon \Forall n > n_0.$

The real number $$x$$ is called the limit of the sequence $$\{ x_n \}$$, and we write $$x_n \to x$$ or $$x = \lim_{n \to \infty} x_n$$.

In other words, a sequence of real numbers $$\{ x_n \}$$ converges to some real number $$x$$ if and only if for each $$\epsilon > 0$$, the terms $$x_n$$ are eventually $$\epsilon$$-close to $$x$$.

Example 2.6 (Sequence convergence)

Consider the sequence $$\{ x_n \}$$, where $$x_n = \frac{1}{\sqrt{n}}$$. For a given $$\epsilon > 0$$, choose $$n_0 > \frac{1}{\epsilon^2}$$. Then, for every $$n > n_0$$, we have

$\left |\frac{1}{n^2} - 0 \right | = \frac{1}{n^2} < \epsilon.$

Thus, the sequence converges to 0.

Definition 2.30 (Divergence)

A sequence which doesn’t converge, is said to diverge.

Theorem 2.4 (Sequence Limit Uniqueness)

A sequence of real numbers can have utmost one limit.

Proof. If a sequence diverges, then there is nothing to prove. Otherwise, suppose a sequence $$\{ x_n \}$$ converges to two limits $$x$$ and $$y$$. Thus, for every $$\epsilon > 0$$, there exist $$n_1, n_2 \in \Nat$$ such that $$|x_n - x | < \epsilon \Forall n > n_1$$ and $$| x_n - y | < \epsilon \Forall n > n_2$$. Now, choose $$n_0 = \max (n_1, n_2)$$. Then, by triangle inequality, for every $$n > n_0$$

$0 \leq | x - y | \leq | x - x_n | + | x_n - y | < \epsilon + \epsilon = 2\epsilon.$

Since this is true for all $$\epsilon > 0$$, hence $$x = y$$. Since $$x,y$$ are arbitrarily close, they must be equal.

Recall that the notion of $$\sup$$ and $$\inf$$ was introduced in the Definition 2.5. The same notation can be used for sequences also.

Definition 2.31 (Upper and lower bounds )

Let $$X = \{ x_n \}$$ be a sequence of $$\RR$$.

• An upper bound of $$X$$ is any $$u \in \RR$$ such that $$x_n \leq u \Forall n \in \Nat$$.

• A lower bound of $$X$$ is any $$l \in \RR$$ such that $$x_n \geq l \Forall n \in \Nat$$.

• If $$X$$ has an upper bound it is said to be bounded from above.

• If $$X$$ has a lower bound it is said to be bounded from below.

• If $$X$$ is both bounded from above and below, then $$X$$ is said to be bounded.

• A real number is called a least upper bound or supremum of $$X$$ if it is an upper bound of $$X$$, and it is less than or equal to every other upper bound of $$X$$.

• The least upper bound is denoted by $$\sup(X)$$.

• A real number is called a greatest lower bound or infimum of $$X$$ if it is a lower bound of $$X$$, and it is greater than or equal to every other lower bound of $$X$$.

• The greatest lower bound is denoted by $$\inf(X)$$.

Remark 2.6

Due to the completeness axiom, if a sequence $$\{x_n\}$$ has an upper bound, it has a least upper bound denoted by $$\sup\{ x_n \}$$ and if it has a lower bound, it has a greatest lower bound denoted by $$\inf\{ x_n \}$$.

The notion of upper boundedness and lower boundedness can be subsumed into a single definition.

Definition 2.32 (Boundedness)

A sequence $$\{ x_n \}$$ is said to be bounded if there exists a number $$M > 0$$ such that $$| x_n | \leq M \Forall n \in \Nat$$.

Theorem 2.5

Every convergent sequence is bounded.

Proof. Let $$\{ x_n \}$$ converge to $$x$$. Choosing a particular value of $$\epsilon = 1$$, there exists $$n_0 \in \Nat$$ such that $$| x_n - x | < 1 \Forall n > n_0$$. Thus, $$x_n \in (x - 1, x + 1)$$. This means that

$| x_n | < | x | + 1 \Forall n > n_0.$

Now define $$M = \max \{|x_1|, |x_2|, \dots, |x_{n_0}|, | x | + 1 \}$$. It follows that $$|x_n | \leq M \Forall n \in \Nat$$ as desired.

Definition 2.33 (Unbounded above )

A sequence $$\{ x_n \}$$ is said to be unbounded above if there exists no $$u \in \RR$$ such that $$x_n \leq u \Forall n \in \Nat$$.

In other words, for every $$u \in \RR$$, there exists an $$x_n > u$$.

Definition 2.34 (Unbounded below )

A sequence $$\{ x_n \}$$ is said to be unbounded below if there exists no $$l \in \RR$$ such that $$x_n \geq l \Forall n \in \Nat$$.

In other words, for every $$l \in \RR$$, there exists an $$x_n < l$$.

## 2.3.1. Monotone Sequences#

Definition 2.35 (Monotone sequences)

• A sequence $$\{ x_n \}$$ is said to be increasing if $$x_n \leq x_{n + 1}$$ for each $$n$$.

• A sequence $$\{ x_n \}$$ is said to be decreasing if $$x_n \geq x_{n + 1}$$ for each $$n$$.

• A sequence $$\{ x_n \}$$ is said to be monotone if it is either increasing or decreasing.

• The notation $$x_n \uparrow x$$ means $$\{ x_n \}$$ is increasing and $$x = \sup \{ x_n \}$$. It applies if $$\{x_n \}$$ is bounded from above.

• The notation $$x_n \downarrow x$$ means $$\{ x_n \}$$ is decreasing and $$x = \inf \{ x_n \}$$. It applies if $$\{x_n \}$$ is bounded from below.

• If a sequence $$\{ x_n \}$$ satisfies $$x_n = c$$ for all $$n$$, then it is called a constant sequence.

An increasing sequence is bounded from below. Its greatest lower bound is $$x_1$$.

A decreasing sequence is bounded from above. Its least upper bound is $$x_1$$.

Remark 2.7 (Unbounded increasing sequences)

Let $$\{x_n\}$$ be an increasing and unbounded sequence. Then for every $$M > 0$$, there exists $$n_0 \in \Nat$$ such that for every $$n > n_0$$, $$x_n > M$$.

Remark 2.8 (Unbounded decreasing sequences)

Let $$\{x_n\}$$ be an decreasing and unbounded sequence. Then for every $$M < 0$$, there exists $$n_0 \in \Nat$$ such that for every $$n > n_0$$, $$x_n < M$$.

Theorem 2.6 (Convergence of bounded monotone sequences)

Every monotone bounded sequence of real numbers is convergent.

Proof. Let $$\{ x_n \}$$ be increasing and bounded sequence. From completeness axiom it follows that there exists $$x = \sup \{ x_n \}$$. We claim that $$x$$ itself is the limit of $$\{ x_n \}$$. From Proposition 2.6 we recall that for every $$\epsilon > 0$$, there exists a number $$x_{n_0} \in \{ x_n \}$$, such that

$x - \epsilon < x_{n_0} \leq x.$

Since $$\{ x_n \}$$ is increasing, hence

$x - \epsilon < x_{n} \leq x \quad \Forall n \geq n_0.$

This means that $$| x - x_n | = x - x_n < \epsilon \Forall n \geq n_0$$. Thus $$x$$ is indeed the limit. We follow similar steps to prove for decreasing sequence.

Theorem 2.7 (Convergence of constant sequences)

Let $$\{ x_n \}$$ be a constant sequence with $$x_n = c$$. Then $$\lim \{ x_n \} = c$$.

Proof. For all $$\epsilon > 0$$, $$| x_n - c | = 0 < \epsilon$$ for all $$n \in \Nat$$.

## 2.3.2. The Calculus of Limits#

Let $$\{ x_n \}$$ and $$\{ y_n \}$$ be convergent sequences of $$\RR$$. Our concern here is to understand what happens to the limits if the sequences are combined.

Let $$\lim \{x_n\} = x$$ and $$\lim \{y_n\} = y$$. Then:

Theorem 2.8 (Scaling a sequence)

$\lim \{\alpha x_n \} = \alpha x \Forall \alpha \in \RR.$

Proof. If $$\alpha = 0$$, then we have a constant sequence and the result is trivial. So assume that $$\alpha \neq 0$$. Then:

$|\alpha x_n - \alpha x | = | \alpha | | x_n - x |.$

Let $$\epsilon > 0$$ and choose $$n_0 \in \Nat$$ such that $$| x - x_n | < \frac{\epsilon}{ | \alpha | }$$ for all $$n > n_0$$. Then

$|\alpha x_n - \alpha x | = | \alpha | | x_n - x | < | \alpha | \frac{\epsilon}{ | \alpha | } = \epsilon \Forall n > n_0.$

Corollary 2.4 (Negating a sequence)

$\lim \{-x_n \} = -x.$

We get this result by choosing $$\alpha = -1$$.

$\lim \{x_n + y_n\} = x + y.$

Proof. From triangle inequality we get:

$| x_n + y_n - (x + y) | \leq | x_n - x | + | y_n - y |.$

For any $$\epsilon > 0$$, choose $$n_1$$ such that $$| x_n - x | < \frac{\epsilon}{2} \Forall n > n_1$$. Similarly, choose $$n_2$$ such that $$| y_n - y | < \frac{\epsilon}{2} \Forall n > n_2$$. Now choose $$n_0 = \max (n_1, n_2)$$. Then:

$| x_n + y_n - (x + y) | \leq | x_n - x | + | y_n - y | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \Forall n > n_0.$

Corollary 2.5 (Subtraction of sequences)

$\lim \{x_n - y_n\} = x - y.$

Negate $$\{ y_n \}$$ and add to $$\{x_n \}$$.

Theorem 2.10 (Multiplication of sequences)

$\lim \{x_n y_n\} = x y.$

Proof. First let us assume that $$x \neq 0$$. We note that:

$\begin{split} | x_n y_n - x y | &= | x_n y_n - x y_n + x y_n - x y | \\ &\leq | x_n y_n - x y_n | + | x y_n - x y |\\ &= | y_n | |x_n - x | + | x | | y_n - y |. \end{split}$

Let $$\epsilon > 0$$ be arbitrary. Choose $$n_1 > 0$$ such that

$n > n_1 \implies |y_n - y | < \frac{1}{| x |} \frac{\epsilon}{2}.$

Since every convergent sequence is bounded, let $$M > 0$$ be a bound of $$\{y_n \}$$ (i.e. $$-M \leq y_n \leq M$$). Choose $$n_2 > 0$$ such that

$n > n_2 \implies |x_n - x | < \frac{1}{M} \frac{\epsilon}{2}.$

Further choose $$n_0 = \max(n_1, n_2)$$. Then, we have

$| x_n y_n - x y | \leq | y_n | |x_n - x | + | x | | y_n - y | < | y_n | \frac{1}{M} \frac{\epsilon}{2} + | x | \frac{1}{| x |} \frac{\epsilon}{2} \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

Since $$\epsilon$$ is arbitrary, hence we have shown that $$\lim \{ x_n y_n \} = x y$$.

Now consider the case where $$x = 0$$. We need to show that $$\lim \{ x_n y_n \} = 0$$. Let $$\epsilon > 0$$ and choose $$n_0$$ such that $$|x_n - 0 | = | x_n | < \frac{\epsilon}{M}$$ for all $$n > n_0$$. Then

$| x_n y_n - 0 | \leq |x_n | | y_n| \leq |x_n | M <\frac{\epsilon}{M} M = \epsilon.$

Theorem 2.11 (Division of sequences)

$\lim \{x_n / y_n\} = x / y \text{ provided } y \neq 0.$

Proof. If we can prove that $$y_n \to y$$ implies that $$\frac{1}{y_n} \to \frac{1}{y}$$ whenever $$y \neq 0$$, then the division of sequences reduces to multiplication of sequences $$\{ x_n \}$$ and $$\{ \frac{1}{y_n} \}$$. But

$\left | \frac{1}{y_n} - \frac{1}{y} \right | = \frac{| y - y_n | }{ | y | |y_n |}.$

Choose $$\epsilon_0 = | y | / 2$$. Then there exists $$n_1 \in \Nat$$ such that $$| y - y_n | < \epsilon_0 = | y | / 2$$ whenever $$n > n_1$$. This gives us, $$| y_n | > | y | / 2$$ whenever $$n > n_1$$ (i.e. $$y_n$$ is so close to $$y$$ that its magnitude is much larger than $$| y | / 2$$). Equivalently, we have

$\frac{1}{ | y_n | } < \frac{2}{| y |} \quad \forall n > n_1.$

Next, for arbitrary $$\epsilon > 0$$, we choose $$n_2$$ such that for all $$n > n_2$$

$| y_n - y | < \frac{\epsilon | y |^2}{2}.$

Finally, pick $$n_0 = \max(n_1, n_2)$$. Then $$n > n_0$$ implies

$\left | \frac{1}{y_n} - \frac{1}{y} \right | = \frac{| y - y_n |}{ | y | |y_n |} < \frac{\epsilon | y |^2}{2} \frac{1}{| y |} \frac{2}{| y |} = \epsilon.$

Thus, $$y_n \to y$$ implies that $$\frac{1}{y_n} \to \frac{1}{y}$$ whenever $$y \neq 0$$. Now division reduces to multiplication and we are done.

Although some elements of $$\{ y_n\}$$ may be zero, but eventually $$y_n$$ becomes arbitrarily close to $$y$$ and since $$y \neq 0$$, hence,

$0 < \frac{|y|}{2} < | y_n | < | y| \Forall n > \text{ some } m.$

In other words, there comes a point $$m$$ in the sequence $$Y$$ so that all elements in $$Y$$ after $$y_m$$ are non-zero with magnitude larger than $$|y|/2$$. We can practically throw away the first $$m$$ terms from both the sequences $$X$$ and $$Y$$ and focus on the convergence of remaining sequence.

Next, we examine some of the order properties of the limits of sequences.

Theorem 2.12 (Order limit theorem)

Assume $$\lim x_n = x$$ and $$\lim y_n = y$$.

1. If $$x_n \geq 0$$ for all $$n \in \Nat$$, then $$x \geq 0$$.

2. If $$x_n \leq y_n$$ for all $$n \in \Nat$$, then $$x \leq y$$.

3. If there exists $$\alpha \in \RR$$ for which $$\alpha \leq y_n$$ for all $$n \in \Nat$$, then $$\alpha \leq y$$. Similarly if $$x_n \leq \alpha$$ for all $$n \in \Nat$$, then $$x \leq \alpha$$.

In words,

1. If a sequence is non-negative, then its limit is non-negative.

2. If one sequence is less than equal to another sequence for every term in the sequence, then its limit is also less than equal to the other sequence.

3. A lower bound of a sequence is less than equal to its limit. An upper bound of a sequence is greater than equal to its limit.

Proof. (1) By contradiction, assume that $$x < 0$$. Then $$x + | x | = 0$$. Now consider $$\epsilon = | x |$$. Since $$\{x_n \}$$ is convergent, there exists $$n_0 \in \Nat$$ such that $$| x_n - x | < \epsilon = | x |$$. Thus,

$| x_n - x | < | x | \implies x - | x | < x_n < x + | x | \implies x_n < 0.$

This is a contradiction. Hence $$x \geq 0$$.

(2) By Corollary 2.5 we have, $$\lim (y_n - x_n) = y - x$$. Since $$y_n \geq x_n$$, hence $$y_n - x_n \geq 0$$. From (1), we get $$y - x \geq 0$$. This implies $$y \geq x$$.

(3) Take $$x_n = \alpha$$. Then $$y_n \geq \alpha = x_n \implies y \geq \alpha$$ (using (2)).

Corollary 2.6 (Order limit theorem extension)

Assume $$\lim x_n = x$$ and $$\lim y_n = y$$.

1. If $$x_n \geq 0$$ for all $$n > n_0$$, then $$x \geq 0$$.

2. If $$x_n \leq y_n$$ for all $$n > n_0$$, then $$x \leq y$$.

3. If there exists $$\alpha \in \RR$$ for which $$\alpha \leq y_n$$ for all $$n > n_0$$, then $$\alpha \leq y$$. Similarly if $$x_n \leq \alpha$$ for all $$n > n_0$$, then $$x \leq \alpha$$.

We throw away the first $$n_0$$ terms from each sequence and apply the theorem on the remaining part(s).

Example 2.7 (Limits don’t preserve strict inequality)

Consider $$x_n = \frac{1}{n}$$ and $$y_n = \frac{1}{n+1}$$. $$\lim x_n = 0$$. $$\lim y_n = 0$$.

Thus, $$x_n > y_n$$ doesn’t imply $$\lim x_n > \lim y_n$$. We only have $$x_n > y_n \implies x_n \geq y_n \implies \lim x_n \geq \lim y_n$$.

Similarly, $$x_n > 0$$ implies that $$\lim x_n \geq 0$$. Or $$x_n < r$$ implies that $$\lim x_n \leq r$$.

Theorem 2.13 (Squeeze theorem for sequences)

If $$x_n \leq y_n \leq z_n$$ for all $$n \in \Nat$$ and if $$\lim x_n = \lim z_n = l$$, then $$\lim y_n = l$$.

Proof. Let $$\lim y_n = y$$. Using order limit theorem, $$x_n \leq y_n$$ gives us $$l \leq y$$ and $$y_n \leq z_n$$ gives us $$y \leq l$$. Thus, $$l \leq y \leq l \implies y = l$$.

Corollary 2.7 (Squeeze theorem for sequences extension)

Let $$\lim x_n = \lim z_n = l$$. If there exists $$n_0 \in \Nat$$ such that for all $$n > n_0$$, $$x_n \leq y_n \leq z_n$$, then $$\lim y_n = l$$.

Drop the first $$n_0$$ terms from all the three sequences and then apply the theorem on remaining sequences.

Theorem 2.14 (Convergence of absolute sequence)

If $$x_n \to x$$, then $$| x_n | \to x$$. But the converse is not true.

Proof. Since $$x_n \to x$$, for every $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that $$n > 0$$ implies $$| x_n - x | < \epsilon$$. By triangle inequality

$| | x_n | - | x | | \leq | x_n - x | < \epsilon.$

This completes the proof.

Now consider the sequence

$\{ 1, -1, 1, -1, 1, -1, \dots \}.$

Although in absolute value it converges to $$1$$, the sequence itself doesn’t converge. Thus, the converse is not true.

## 2.3.3. Infinite Series#

Definition 2.36

Let $$\{ x_n \}$$ be a sequence. An infinite series is a formal expression of the form

$\sum_{n = 1}^{\infty} x_n = x_1 + x_2 + x_3 + x_4 + \dots.$

The corresponding sequence of partial sums $$\{ s_m\}$$ is defined as

$s_m = x_1 + \dots + x_m.$

We say that the series $$\sum_{n = 1}^{\infty} x_n$$ converges to some $$s \in \RR$$ if the sequence $$\{ s_m \}$$ converges to $$s$$. In this case we write

$\sum_{n = 1}^{\infty} x_n = s.$

Example 2.8 (Convergent series)

Consider

$\sum_{n = 1}^{\infty} \frac{1}{n^2}.$

Looking at the partial sums, we observe:

$\begin{split} s_m &= 1 + \frac{1}{4} + \frac{1}{9} \dots + \frac{1}{m^2}\\ & < 1 + \frac{1}{2 \cdot 1} + \frac{1}{3 \cdot 2} + \dots \frac{1}{m \cdot (m - 1)}\\ & = 1 + \left (1 - \frac{1}{2} \right) + \left (\frac{1}{2} - \frac{1}{3} \right) + \dots + \left (\frac{1}{m - 1} - \frac{1}{m} \right) \\ & = 1 + 1 - \frac{1}{m} \\ &< 2. \end{split}$

Thus, $$2$$ is an upper bound of the sequence of partial sums. Hence, by monotone convergence theorem, the series converges to some (unknown) limit less than 2.

Example 2.9 (Harmonic series)

Consider

$\sum_{n = 1}^{\infty} \frac{1}{n}.$

We note that

$s_4 = 1 + \frac{1}{2} + \left (\frac{1}{3} + \frac{1}{4} \right ) > 1 + \frac{1}{2} + \left (\frac{1}{4} + \frac{1}{4} \right ) = 2.$

Similarly, we find that $$s_8 > 2\frac{1}{2}$$. Further, we note that:

$\begin{split} s_{2^k} &= 1 + \frac{1}{2} + \left (\frac{1}{3} + \frac{1}{4} \right ) + \left (\frac{1}{5} + \dots + \frac{1}{8} \right ) + \dots + \left (\frac{1}{2^{k - 1} + 1} + \dots + \frac{1}{2^k} \right ) \\ &> 1 + \frac{1}{2} + \left (\frac{1}{4} + \frac{1}{4} \right ) + \left (\frac{1}{8} + \dots + \frac{1}{8} \right ) + \dots + \left (\frac{1}{2^k} + \dots + \frac{1}{2^k} \right ) \\ &= 1 + \frac{1}{2} + 2 \frac{1}{4} + 4 \frac{1}{8} + \dots + 2^{k - 1} \frac{1}{2^k} \\ &= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots + \frac{1}{2}\\ &= 1 + k \frac{1}{2}. \end{split}$

Thus, the harmonic series is unbounded.

Theorem 2.15 (Cauchy condensation test)

Suppose $$\{ x_n \}$$ is decreasing and satisfies $$x_n \geq 0$$ for all $$n \in \Nat$$. Then, the series $$\sum_{n = 1}^{\infty} x_n$$ converges if and only if the series

$\sum_{n = 0}^{\infty} 2^n x_{2^n} = x_1 + 2 x_2 + 4 x_4 + 8 x_8 + 16 x_{16} + \dots$

converges.

Proof. Let $$y_n = 2^{n-1} x_{2^{n-1}}$$. Then, the second series is $$\sum_{n = 1}^{\infty} y_n$$. Let the partial sums of $$\{ x_n \}$$ be $$s_m$$ and the partial sums of $$\{ y_n\}$$ be $$t_k$$.

First, assume that $$\sum_{n = 1}^{\infty} y_n$$ converges. Thus, the sequence $$\{ t_k \}$$ converges. Since every convergent sequence is bounded, the sequence $$\{ t_k \}$$ is bounded. Thus, there exists $$M > 0$$ such that $$t_k \leq M$$ for all $$k \in \Nat$$. Since $$x_n \geq 0$$, the partial sums $$s_m$$ are increasing. Thus, if we show that $$\{ s_m \}$$ is bounded, then by monotone convergence theorem, we would have shown that $$\{s_m\}$$ converges, hence the series $$\sum_{n = 1}^{\infty} x_n$$ converges.

Let us fix $$m$$ and choose $$k$$ to be large enough so that $$m \leq 2^{k + 1} - 1$$. Then $$s_m \leq s_{2^{k + 1} - 1}$$. Now,

$\begin{split} s_{2^{k + 1} - 1} &= x_1 + (x_2 + x_3) + (x_4 + \dots x_7) + \dots + \dots + (x_{2^k} + \dots + x_{2^{k + 1} - 1})\\ &\leq x_1 + (x_2 + x_2) + (x_4 + x_4 + x_4 + x_4) + \dots + (x_{2^k} + \dots + x_{2^k})\\ &= x_1 + 2 x_2 + \dots + 2^k x_{2^k}\\ &= y_1 + y_2 + \dots + y_k \\ &= t_k. \end{split}$

Thus, $$s_m \leq t_k \leq M$$. $$\{s_m \}$$ is bounded, hence convergent.

We now show that if $$\sum_{n = 1}^{\infty} y_n$$ diverges, then $$\sum_{n = 1}^{\infty} x_n$$ diverges too.

Consider the sum

$\begin{split} s_{2^k} &= x_1 + x_2 + (x_3 + x_4) + (x_5 + \dots + x_8) + \dots + (x_{2^{k -1 } + 1} + \dots x^{2^k})\\ &\geq x_1 + x_2 + (x_4 + x_4) + (x_8 + \dots + x_8) + \dots + (x_{2^k} + \dots x_{2^k})\\ &= x_1 + x_2 + 2 x_4 + 4 x_8 + \dots 2^{k - 1} x_{2^k}\\ &= \frac{1}{2} x_1 + \frac{1}{2} \left (x_1 + 2 x_2 + 4 x_4 + 8 x_8 + \dots + 2^k x^{2^k} \right )\\ &= \frac{1}{2} x_1 + t_k \geq t_k. \end{split}$

Now, since $$\{ t_k \}$$ diverges, hence $$\{ s_{2^k} \}$$ too diverges. Thus, the series diverges.

Definition 2.37 (Absolutely summable)

A series $$\sum x_n$$ is called absolutely summable if $$\sum |x_n|$$ converges.

A sequence $$\{x_n \}$$ is called absolutely summable if $$\sum |x_n|$$ converges.

## 2.3.4. Subsequences#

Theorem 2.16 (Subsequence convergence)

Subsequences of a convergent sequence converge to the same limit as the original sequence. If $$\lim_{n \to \infty} x_n = x$$, then $$\lim_{n \to \infty} y_n = x$$ for every subsequence $$\{ y_n \}$$ of $$\{ x_n \}$$.

Conversely, if two different subsequences of $$\{ x_n \}$$ converge to different limits, then the sequence $$\{ x_n \}$$ does not converge.

Proof. Since $$\lim_{n \to \infty} x_n = x$$, for every $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that $$| x - x_n | < \epsilon \Forall n > n_0$$. Now, if $$\{ y_n \}$$ is a subsequence, then there exists a strictly increasing sequence $$\{ k_n \}$$ of natural numbers (i.e. $$1 \leq k_1 < k_2 < k_3 < \ldots)$$ such that $$y_n = x_{k_n}$$ holds for each $$n$$. Clearly, there exists a $$k_0 > 0$$ such that $$k_n \geq n_0 \Forall n > k_0$$. Then, $$| x - y_n | < \epsilon \Forall n > k_0$$. Thus, $$\{ y_n \}$$ converges to $$x$$ too.

Theorem 2.17 (Bolzano Weierstrass theorem)

Every bounded sequence contains a convergent subsequence.

The proof of this theorem follows a constructive approach (i.e., we will construct a subsequence and show that it is convergent). We construct a sequence of nested closed intervals with increasingly smaller lengths and pick a point in each such interval to form a subsequence.

Proof. Let $$\{ x_n \}$$ be a bounded sequence. Thus, there exists $$M > 0$$ such that $$| x_n | \leq M$$ for all $$n \in \Nat$$. Divide the interval $$[-M, M]$$ into two equal closed intervals $$[-M, 0]$$ and $$[0, M]$$. At least one of the two halves must have an infinite number of points in $$\{ x_n \}$$ (since if both halves had finite number of points, then the total number of points in the sequence would be finite which is a contradiction). Choose a half for which this is true, and label this half as $$I_1$$. Choose a point $$x_{n_1} \in I_1$$. Now divide $$I_1$$ into two equal closed intervals. Again, since $$I_1$$ contains infinite number of points, hence at least one of the halves must have infinite number of points. Pick a half which contains infinite number of points and label it as $$I_2$$. Now, pick a point $$x_{n_2}$$ from $$I_2$$ such that $$n_2 > n_1$$. In general, construct a closed interval $$I_k$$ from a half of $$I_{k-1}$$ containing infinite number of points. Further, we choose a point $$x_{n_k}$$ such that $$n_k > n_{k-1} > \dots > n_2 > n_1$$ and $$x_{n_k} \in I_k$$. We claim that the subsequence $$\{x_{n_k} \}$$ is a convergent subsequence. For this, we need a limit for the sequence. Since

$I_1 \supseteq I_2 \supseteq \dots \supseteq I_k \supseteq \dots$

are a nested sequence of closed intervals, hence by nested interval property, their intersection $$I = \bigcap_{k=1}^{\infty} I_k$$ is non-empty. Actually, it’s easy to show that this $$I$$ is a singleton too. If there were two distinct points $$x, y$$ in $$I$$, then considering $$d = | x - y | > 0$$, we could find an interval $$I_j$$ whose length is smaller than $$d$$. Thus both $$x, y$$ could not fit in $$I_j$$. Hence $$I$$ contains only one point. Let $$I = \{ x \}$$. We now show that $$x_{n_k} \to x$$.

Let $$\epsilon > 0$$. By construction, the length of $$I_k$$ is $$M\frac{1}{2^{k-1}}$$. Since it converges to 0, hence, we can choose $$n_0 \in \Nat$$ such that for every $$k > n_0$$, the length of $$I_k$$ is less than $$\epsilon$$. Since $$x$$ and $$x_{n_k}$$ are both in $$I_k$$, hence it follows that $$| x_{n_k} - x| < \epsilon$$.

## 2.3.5. Cauchy Sequence#

Definition 2.38 (Cauchy sequence)

A sequence $$\{ x_n \}$$ in $$\RR$$ is called a Cauchy sequence if, for every $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ (depending on $$\epsilon$$) such that whenever $$m, n > n_0$$ it follows that $$| x_m - x_n | < \epsilon$$.

Remark 2.9

A little thought would show that saying $$m, n > n_0$$ or $$m, n \geq n_1$$ doesn’t make much difference in the definition. The two thresholds can be related by : $$n_1 = n_0 + 1$$.

Theorem 2.18 (Boundedness of Cauchy sequences)

A Cauchy sequence is bounded.

Proof. Let $$\{ x_n \}$$. Choose $$\epsilon = 1$$. Then there exists $$n_0 \in \Nat$$ such that $$| x_n - x_m | < 1$$ whenever $$m, n \geq n_0$$. In particular, the statement is valid when $$m = n_0$$. i.e. $$| x_n - x_{n_0} | < 1$$ . But,

$| x_n - x_{n_0} | < 1 \implies | | x_n | - | x_{n_0 } | | < 1 \implies |x_n | < 1 + | x_{n_0 } | \Forall n \geq n_0.$

Choosing $$M = \max(|x_1|, \dots, |x_{n_0-1}|, |x_{n_0}| + 1)$$, it is clear that $$| x_n | \leq M$$, hence $$\{ x_n \}$$ is bounded.

Theorem 2.19 (Convergence of Cauchy sequences)

A Cauchy sequence is convergent.

Proof. Let $$\{ x_n \}$$ be a Cauchy sequence. Hence it is bounded. Hence, by Bolzano Weierstrass theorem, it has a convergent subsequence.

Let $$\{ x_{n_k} \}$$ be such a convergent subsequence with the limit $$\lim x_{n_k} = x$$.

Thus, for any $$\epsilon > 0$$, there exists $$n_1 \in \Nat$$ such that for all $$k > n_1$$,

$| x_{n_k} - x | < \frac{\epsilon}{2}.$

Also, since $$\{x_n \}$$ is Cauchy, there exists $$n_2 \in \Nat$$ such that for all $$n, m > n_2$$,

$|x_n - x_m | < \frac{\epsilon}{2}.$

Pick any $$k > n_1$$ such that $$n_k > n_2$$. Then, for all $$n > n_2$$,

$|x_n - x | \leq |x_n - x_{n_k} | + |x_{n_k} - x | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

Thus, $$\lim x_n = x$$.

Theorem 2.20 (Convergence and Cauchyness)

A sequence of real numbers converges if and only if it is a Cauchy sequence.

Proof. Let a sequence $$\{ x_n \}$$ converge to $$\epsilon$$. Then for every $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that $$| x_n - x | < \epsilon / 2$$ for all $$n > n_0$$. Clearly, if $$m, n > n_0$$, then

$| x_m - x_n | \leq | x_m - x | + | x - x_n | < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

Thus, $$\{ x_n \}$$ is Cauchy.

For the converse, in the previous theorem, we proved that a Cauchy sequence is convergent.

Remark 2.10

Let $$\{ x_n \}$$ be a Cauchy sequence with $$\lim x_n = x$$. Let $$\epsilon > 0$$. Choose $$n_0$$ such that $$m, n > n_0$$ implies $$|x_m - x_n| < \epsilon$$.

Then

$|x_n - x | \leq \epsilon \Forall n > n_0.$

Proof. We have, for all $$m, n > n_0$$

$|x_m - x_n | < \epsilon \iff x_m - \epsilon < x_n < x_m + \epsilon.$

We will fix $$n$$ and vary $$n$$ to compute the limit inequalities.

1. Consider the strict inequality: $$x_m - \epsilon < x_n$$ for all $$m > n_0$$.

2. Taking the limit on the sequence $$x_m$$ (in L.H.S.), we get: $$x - \epsilon \leq x_n$$.

3. Consider the strict inequality: $$x_n < x_m + \epsilon$$ for all $$m > n_0$$.

4. Taking the limit on the sequence $$x_m$$ (in R.H.S.), we get: $$x_n \leq x + \epsilon$$.

5. Together, we get: $$x - \epsilon \leq x_n \leq x + \epsilon$$.

6. Combining, we get $$|x_n -x| \leq \epsilon$$ for all $$n > n_0$$.

## 2.3.6. Limit Inferior and Limit Superior#

Theorem 2.21 (Sequences of partial suprema and infima)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Define

$s_n = \sup \{ x_k \ST k \geq n\}$

and

$t_n = \inf \{ x_k \ST k \geq n\}.$

If $$\{ x_n \}$$ is not bounded above, then

$\lim_{n \to \infty} s_n = \infty.$

If $$\{ x_n \}$$ is not bounded below, then

$\lim_{n \to \infty} t_n = -\infty.$

If $$\{x_n\}$$ is bounded from above, then $$\{ s_n \}$$ is a nonincreasing sequence.

If $$\{x_n\}$$ is bounded from below, then $$\{ t_n \}$$ is a nondecreasing sequence.

If $$\{ x_n \}$$ is bounded, then both the sequences $$\{ s_n \}$$ and $$\{ t_n \}$$ are convergent.

Proof. Let $$\{x_n \}$$ not be bounded from above.

1. Then, for any $$n \in \Nat$$, the set $$\{ x_k \ST k \geq n \}$$ is also not bounded from above.

2. Thus, $$s_n = \sup \{ x_k \ST k \geq n\} = \infty$$ for all $$n \in \Nat$$.

3. Thus, $$\lim_{n \to \infty} s_n = \infty$$.

Let $$\{x_n \}$$ not be bounded from below.

1. Then, for any $$n \in \Nat$$, the set $$\{ x_k \ST k \geq n \}$$ is also not bounded from below.

2. Thus, $$t_n = \inf \{ x_k \ST k \geq n\} = -\infty$$ for all $$n \in \Nat$$.

3. Thus, $$\lim_{n \to \infty} t_n = -\infty$$.

We note that for $$m < n$$,

$\{ x_k \ST k \geq n\} \subseteq \{ x_k \ST k \geq m\}.$

Now, assume that $$\{x_n \}$$ is bounded from above.

1. Then, $$\sup \{ x_k \ST k \geq n\} \leq \sup \{ x_k \ST k \geq m\}$$.

2. Or $$s_n \leq s_m$$.

3. Thus, $$m < n$$ implies that $$s_m \geq s_n$$.

4. Thus, $$\{ s_n \}$$ is a nonincreasing sequence.

Now, assume that $$\{x_n \}$$ is bounded from below.

1. Then, $$\inf \{ x_k \ST k \geq n\} \geq \inf \{ x_k \ST k \geq m\}$$.

2. Or $$t_n \geq t_m$$.

3. Thus, $$m < n$$ implies that $$t_m \leq t_n$$.

4. Thus, $$\{ t_n \}$$ is a nondecreasing sequence.

Now, assume that $$\{x_n \}$$ is bounded.

1. From Theorem 2.6, a monotone bounded sequence is convergent.

2. Since $$\{ x_n \}$$ is bounded, hence $$\{ s_n \}$$ is bounded too.

3. Since $$\{ x_n \}$$ is bounded, hence $$\{ t_n \}$$ is bounded too.

4. Thus, both $$\{ s_n \}$$ and $$\{ t_n \}$$ are bounded and monotone.

5. Thus, both of them are convergent sequences.

Definition 2.39 (Limit superior and inferior)

Let $$\{x_n \}$$ be a sequence of $$\RR$$. The limit superior of the sequence is defined as:

$\limsup_{n \to \infty} x_n \triangleq \lim_{n \to \infty} \sup \{x_k \ST k \geq n \}.$

Similarly, the limit inferior of the sequence is defined as:

$\liminf_{n \to \infty} x_n \triangleq \lim_{n \to \infty} \inf \{x_k \ST k \geq n \}.$

If we define

$s_n = \sup \{ x_k \ST k \geq n\} \text{ and } t_n = \inf \{ x_k \ST k \geq n\}$

then,

$\limsup_{n \to \infty} x_n = \lim_{n \to \infty} s_n$

and

$\liminf_{n \to \infty} x_n = \lim_{n \to \infty} t_n.$

It is imperative to establish that the definition of limit inferior and limit superior is justified.

1. If $$\{x_n \}$$ is not bounded from above, then $$s_n = \infty$$.

2. If $$\{x_n \}$$ is bounded from above then $$\{ s_n \}$$ is nondecreasing.

3. In this case, if $$\{x_n \}$$ is bounded from below, then $$\{ s_n \}$$ converges, otherwise it diverges to $$-\infty$$.

Similar justification applies for the limit inferior too.

Remark 2.11 (Limit superior and inferior for unbounded sequences)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$.

If $$\{ x_n \}$$ is not bounded from above, then

$\limsup_{n \to \infty} x_n = \infty.$

If $$\{ x_n \}$$ is not bounded from below, then

$\liminf_{n \to \infty} x_n = -\infty.$

Theorem 2.22 (Limit superior $$\geq$$ limit inferior)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Then,

$\liminf_{n \to \infty} x_n \leq \limsup_{n \to \infty} x_n.$

Proof. If we define

$s_n = \sup \{ x_k \ST k \geq n\} \text{ and } t_n = \inf \{ x_k \ST k \geq n\}$

then, $$t_n \leq s_n$$ for every $$n$$.

Then, by Theorem 2.12,

$\lim_{n \to \infty} t_n \leq \lim_{n \to \infty} s_n.$

Thus,

$\liminf_{n \to \infty} x_n = \lim_{n \to \infty} t_n \leq \lim_{n \to \infty} s_n = \limsup_{n \to \infty} x_n.$

Theorem 2.23 (Relationship between limit superior and inferior)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Then,

$\limsup_{n \to \infty} (-x_n) = - \liminf_{n \to \infty} x_n.$

Proof. We recall that

$\inf(-1 \cdot A) = -1 \cdot \sup (A) \text{ and } \sup(-1 \cdot A) = -1 \cdot \inf \sup (A)$

Thus,

$\begin{split} \limsup_{n \to \infty} (-x_n) &= \lim_{n \to \infty} \sup \{-x_k \ST k \geq n \}\\ &= \lim_{n \to \infty} -\inf \{x_k \ST k \geq n \}\\ &= - \lim_{n \to \infty} \inf \{x_k \ST k \geq n \}\\ &= - \liminf_{n \to \infty} x_n. \end{split}$

Theorem 2.24 (Characterization of limit superior)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Let $$u \in \RR$$. The following are equivalent.

1. $$\limsup_{n \to \infty} x_n = u$$.

2. For any $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that

$x_n < u + \epsilon \Forall n \geq n_0.$

and there exists a subsequence $$\{x_{k_n} \}$$ of $$\{ x_n \}$$ such that $$\lim_{n \to \infty} x_{k_n} = u$$.

Proof. TBD

Theorem 2.25 (Characterization of limit inferior)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Let $$l \in \RR$$. The following are equivalent.

1. $$\liminf_{n \to \infty} x_n = l$$.

2. For any $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that

$x_n > l - \epsilon \Forall n \geq n_0.$

and there exists a subsequence $$\{x_{k_n} \}$$ of $$\{ x_n \}$$ such that $$\lim_{n \to \infty} x_{k_n} = l$$.

Proof. TBD

This leads us to the fact that the limit of a sequence exists if and only if its limit inferior and limit superior are identical.

### 2.3.6.1. Existence of Limit#

Theorem 2.26 (Limit = limit superior = limit inferior)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Then,

$\lim_{n \to \infty} x_n = l \text{ if and only if } \limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n = l.$

In other words the limit of a sequence exists if and only if both limit superior and limit inferior are equal and in this case, the limit of sequence equals the limit superior and inferior.

Proof. TBD

### 2.3.6.2. Subsequences#

Theorem 2.27 (Convergent subsequences and limit superior/inferior)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$. Let $$\{ x_{k_n} \}$$ be an arbitrary subsequence of $$\{ x_n \}$$.

Suppose $$\limsup x_n = u$$. If $$\{ x_{k_n} \}$$ converges then

$\lim_{n \to \infty} x_{k_n} \leq u.$

Suppose $$\liminf x_n = l$$. If $$\{ x_{k_n} \}$$ converges then

$\lim_{n \to \infty} x_{k_n} \geq l.$

Proof. Assume that $$\limsup x_n = u$$.

1. Let $$\epsilon > 0$$.

2. By Theorem 2.24, there exists $$n_0$$ such that for all $$n > n_0$$

$x_n < u + \epsilon.$
3. Let $$\lim_{n \to \infty} x_{k_n} = s$$.

4. Then, there exists $$n_1$$ such that for all $$k_n > n_1$$

$s - \epsilon < x_{k_n} < s + \epsilon.$
5. Let $$n_2 = \max(n_0, n_1)$$.

6. Then, for all $$k_n > n_2$$

$s - \epsilon < x_{k_n} < u + \epsilon.$
7. Thus, $$s < u + 2 \epsilon$$ for every $$\epsilon > 0$$.

8. Thus, $$s \leq u$$.

The proof for limit inferior is similar.

Remark 2.12 (The set of subsequential limits for bounded sequences)

Let $$\{ x_n \}$$ be a bounded sequence of $$\RR$$. Define

$A = \{ x \in \RR \ST \text{ there exists a subsequence } \{ x_{k_n} \} \text{ with } \lim x_{k_n} = x \}.$

This is the set of limits of convergent subsequences of $$\{ x_n \}$$. By Theorem 2.17 $$\{ x_n \}$$ has a convergent subsequence since it is bounded. Hence, $$A$$ is not empty.

Each element of $$A$$ is called a subsequential limit of $$\{x_n \}$$. Due to Theorem 2.21, both $$\limsup x_n$$ and $$\liminf x_n$$ are finite since $$\{ x_n \}$$ is bounded.

Let $$a \in A$$. Then, by Theorem 2.27:

$\liminf x_n \leq a \leq \limsup x_n.$

By Theorem 2.24, there exists $$u \in A$$ such that

$\limsup x_n = u.$

By Theorem 2.25, there exists $$l \in A$$ such that

$\liminf x_n = l.$

Thus,

$\limsup x_n = \max A \text{ and } \liminf x_n = \min A.$

### 2.3.6.3. Arithmetic#

Theorem 2.28 (Arithmetic of limit superior and inferior)

Let $$\{ a_n \}$$ and $$\{ b_n \}$$ be sequences of $$\RR$$.

$\limsup_{n \to \infty} (a_n + b_n) \leq \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n.$

$\liminf_{n \to \infty} (a_n + b_n) \geq \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n.$

If both $$\{ a_n \}$$ and $$\{ b_n \}$$ are nonnegative, then

$\limsup_{n \to \infty} (a_n b_n) \leq \left (\limsup_{n \to \infty} a_n \right) \left ( \limsup_{n \to \infty} b_n \right)$

and

$\liminf_{n \to \infty} (a_n b_n) \geq \left (\liminf_{n \to \infty} a_n \right) \left ( \liminf_{n \to \infty} b_n \right).$

### 2.3.6.4. Order#

Theorem 2.29 (Order properties of limit superior and inferior)

Let $$\{ x_n \}$$ and $$\{ y_n \}$$ be sequences of $$\RR$$.

1. If $$x_n \geq 0$$ for all $$n \in \Nat$$, then $$\liminf_{n \to \infty} x \geq 0$$.

2. If $$x_n \leq 0$$ for all $$n \in \Nat$$, then $$\limsup_{n \to \infty} x \leq 0$$.

3. If $$x_n \leq y_n$$ for all $$n \in \Nat$$, then $$\limsup x_n \leq \limsup y_n$$.

4. If $$x_n \geq y_n$$ for all $$n \in \Nat$$, then $$\liminf x_n \geq \liminf y_n$$.

Proof. Assume that $$x_n \geq 0$$.

1. Then, $$\inf \{x_k \ST k \geq n \} \geq 0$$.

2. Thus,

$\liminf_{n \to \infty} x_n = \lim_{n \to \infty} \inf \{x_k \ST k \geq n \} \geq \lim_{n \to \infty} 0 = 0.$

Assume that $$x_n \leq 0$$.

1. Then, $$\sup \{x_k \ST k \geq n \} \leq 0$$.

2. Thus,

$\limsup_{n \to \infty} x_n = \lim_{n \to \infty} \sup \{x_k \ST k \geq n \} \leq \lim_{n \to \infty} 0 = 0.$

Assume that $$x_n \leq y_n$$ for all $$n \in \Nat$$.

1. Choose any $$k \in \Nat$$.

2. Let $$M_n = \sup \{y_k \ST k \geq n \}$$.

3. Then, $$y_k \leq M_n$$ for every $$k \geq n$$.

4. Then, $$x_k \leq y_k \leq M_n$$ for every $$k \geq n$$.

5. Taking supremum over $$k \geq n$$, $$m_n = \sup \{x_k \ST k \geq n \} \leq M_n$$.

6. Thus, $$m_n \leq M_n$$ for every $$n$$.

7. Thus, by Theorem 2.12, $$\lim m_n \leq \lim M_n$$.

8. Thus, $$\limsup x_n \leq \limsup y_n$$.

A similar argument can be used for $$x_n \geq y_n$$ also.

### 2.3.6.5. Damping and Growing Sequences#

Theorem 2.30 (Damping sequences)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$ such that $$x_n > 0$$ for every $$n$$.

Assume that the following holds:

$\limsup_{n \to \infty} \frac{x_{n+1}}{x_n} = u < 1.$

Then, $$\lim_{n \to \infty} x_n = 0$$.

Proof. We proceed as follows.

1. Since $$\frac{x_{n+1}}{x_n} > 0$$, hence $$u \geq 0$$.

2. Since $$u < 1$$, we can choose an $$\epsilon > 0$$ such that $$u + \epsilon < 1$$.

3. Let $$q = u + \epsilon$$. Then, $$0 < q < 1$$.

4. By Theorem 2.24, there exists $$n_0 \in \Nat$$ such that for all $$n \geq n_0$$

$\frac{x_{n+1}}{x_n} < u + \epsilon = q.$
5. Thus, $$x_{n+1} < q x_n$$ for every $$n \geq n_0$$.

6. Let $$x = x_{n_0}$$.

7. Then, we have $$x_n < q^{n - n_0} x$$ for every $$n > n_0$$.

8. Also, $$\lim_{n \to \infty} q^{n - n_0} x = 0$$ since $$q < 1$$.

9. We have $$0 < x_n < q^{n - n_0} x$$ for all $$n > n_0$$.

10. Hence, due to the squeeze theorem, $$\lim_{n \to \infty} x_n = 0$$.

Theorem 2.31 (Growing sequences)

Let $$\{ x_n \}$$ be a sequence of $$\RR$$ such that $$x_n > 0$$ for every $$n$$.

Assume that the following holds:

$\liminf_{n \to \infty} \frac{x_{n+1}}{x_n} = l > 1.$

Then, $$\lim_{n \to \infty} x_n = \infty$$.

Proof. We proceed as follows.

1. Since $$l > 1$$, we can choose an $$\epsilon > 0$$ such that $$l - \epsilon > 1$$.

2. Let $$q = l - \epsilon$$. Then, $$q > 1$$.

3. By Theorem 2.25, there exists $$n_0 \in \Nat$$ such that for all $$n \geq n_0$$

$\frac{x_{n+1}}{x_n} > l - \epsilon = q.$
4. Thus, $$x_{n+1} > q x_n$$ for every $$n \geq n_0$$.

5. Let $$x = x_{n_0}$$.

6. Then, we have $$x_n > q^{n - n_0} x$$ for every $$n > n_0$$.

7. Also, $$\lim_{n \to \infty} q^{n - n_0} x = \infty$$ since $$q > 1$$.

8. We have $$x_n > q^{n - n_0} x$$ for all $$n > n_0$$.

9. Hence, $$\lim x_n \geq \lim q^{n - n_0} x = \infty$$.

10. Thus, $$\lim x_n = \infty$$.