# Sequences and Series

## Contents

# 2.3. Sequences and Series#

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

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.

(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

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 \(x\).

(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

Thus, the sequence converges to 0.

(Divergence)

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

(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 \(n > n_0\)

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.

(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 \(\inf(X)\).

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.

(Boundedness)

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

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

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.

(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 \(x_n > u\).

(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#

(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 \(x_1\).

(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 \(n > n_0\), \(x_n > M\).

(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 \(n > n_0\), \(x_n < M\).

(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

Since \(\{ x_n \}\) is increasing, hence

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.

(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:

(Scaling a sequence)

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

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

(Negating a sequence)

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

(Addition of sequences)

Proof. From triangle inequality we get:

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:

(Subtraction of sequences)

Negate \(\{ y_n \}\) and add to \(\{x_n \}\).

(Multiplication of sequences)

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

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

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

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

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

(Division of sequences)

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

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

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

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

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,

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.

(Order limit theorem)

Assume \(\lim x_n = x\) and \(\lim y_n = y\).

If \(x_n \geq 0\) for all \(n \in \Nat\), then \(x \geq 0\).

If \(x_n \leq y_n\) for all \(n \in \Nat\), then \(x \leq y\).

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,

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

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.

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,

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)).

(Order limit theorem extension)

Assume \(\lim x_n = x\) and \(\lim y_n = y\).

If \(x_n \geq 0\) for all \(n > n_0\), then \(x \geq 0\).

If \(x_n \leq y_n\) for all \(n > n_0\), then \(x \leq y\).

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).

(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 \(\lim x_n \leq r\).

(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, \(l \leq y \leq l \implies y = l\).

(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.

(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

This completes the proof.

Now consider the sequence

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#

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

The corresponding *sequence of partial sums* \(\{ s_m\}\) is defined as

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

(Convergent series)

Consider

Looking at the partial sums, we observe:

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.

(Harmonic series)

Consider

We note that

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

Thus, the harmonic series is unbounded.

(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

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,

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

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

(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#

(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.

(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

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#

(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 \(| x_m - x_n | < \epsilon\).

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 : \(n_1 = n_0 + 1\).

(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,

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.

(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 \(k > n_1\),

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

Pick any \(k > n_1\) such that \(n_k > n_2\). Then, for all \(n > n_2\),

Thus, \(\lim x_n = x\).

(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

Thus, \(\{ x_n \}\) is Cauchy.

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

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

Proof. We have, for all \(m, n > n_0\)

We will fix \(n\) and vary \(n\) to compute the limit inequalities.

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

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

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

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

Together, we get: \(x - \epsilon \leq x_n \leq x + \epsilon\).

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

See also Example 2.7.

## 2.3.6. Limit Inferior and Limit Superior#

(Sequences of partial suprema and infima)

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

and

If \(\{ x_n \}\) is not bounded above, then

If \(\{ x_n \}\) is not bounded below, then

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.

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

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

Thus, \(\lim_{n \to \infty} s_n = \infty\).

Let \(\{x_n \}\) not be bounded from below.

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

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

Thus, \(\lim_{n \to \infty} t_n = -\infty\).

We note that for \(m < n\),

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

Then, \(\sup \{ x_k \ST k \geq n\} \leq \sup \{ x_k \ST k \geq m\}\).

Or \(s_n \leq s_m\).

Thus, \(m < n\) implies that \(s_m \geq s_n\).

Thus, \(\{ s_n \}\) is a nonincreasing sequence.

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

Then, \(\inf \{ x_k \ST k \geq n\} \geq \inf \{ x_k \ST k \geq m\}\).

Or \(t_n \geq t_m\).

Thus, \(m < n\) implies that \(t_m \leq t_n\).

Thus, \(\{ t_n \}\) is a nondecreasing sequence.

Now, assume that \(\{x_n \}\) is bounded.

From Theorem 2.6, a monotone bounded sequence is convergent.

Since \(\{ x_n \}\) is bounded, hence \(\{ s_n \}\) is bounded too.

Since \(\{ x_n \}\) is bounded, hence \(\{ t_n \}\) is bounded too.

Thus, both \(\{ s_n \}\) and \(\{ t_n \}\) are bounded and monotone.

Thus, both of them are convergent sequences.

(Limit superior and inferior)

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

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

If we define

then,

and

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

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

If \(\{x_n \}\) is bounded from above then \(\{ s_n \}\) is nondecreasing.

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.

(Limit superior and inferior for unbounded sequences)

Let \(\{ x_n \}\) be a sequence of \(\RR\).

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

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

\(\geq\) limit inferior)

(Limit superiorLet \(\{ x_n \}\) be a sequence of \(\RR\). Then,

Proof. If we define

then, \(t_n \leq s_n\) for every \(n\).

Then, by Theorem 2.12,

Thus,

(Relationship between limit superior and inferior)

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

Proof. We recall that

Thus,

(Characterization of limit superior)

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

\(\limsup_{n \to \infty} x_n = u\).

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

(Characterization of limit inferior)

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

\(\liminf_{n \to \infty} x_n = l\).

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#

(Limit = limit superior = limit inferior)

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

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#

(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

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

Proof. Assume that \(\limsup x_n = u\).

Let \(\epsilon > 0\).

By Theorem 2.24, there exists \(n_0\) such that for all \(n > n_0\)

\[ x_n < u + \epsilon. \]Let \(\lim_{n \to \infty} x_{k_n} = s\).

Then, there exists \(n_1\) such that for all \(k_n > n_1\)

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

Then, for all \(k_n > n_2\)

\[ s - \epsilon < x_{k_n} < u + \epsilon. \]Thus, \(s < u + 2 \epsilon\) for every \(\epsilon > 0\).

Thus, \(s \leq u\).

The proof for limit inferior is similar.

(The set of subsequential limits for bounded sequences)

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

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:

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

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

Thus,

### 2.3.6.3. Arithmetic#

(Arithmetic of limit superior and inferior)

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

The limit superior satisfies subadditivity.

The limit inferior satisfies superadditivity.

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

and

### 2.3.6.4. Order#

(Order properties of limit superior and inferior)

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

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

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

If \(x_n \leq y_n\) for all \(n \in \Nat\), then \(\limsup x_n \leq \limsup y_n\).

If \(x_n \geq y_n\) for all \(n \in \Nat\), then \(\liminf x_n \geq \liminf y_n\).

Proof. Assume that \(x_n \geq 0\).

Then, \(\inf \{x_k \ST k \geq n \} \geq 0\).

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 \(x_n \leq 0\).

Then, \(\sup \{x_k \ST k \geq n \} \leq 0\).

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 \(n \in \Nat\).

Choose any \(k \in \Nat\).

Let \(M_n = \sup \{y_k \ST k \geq n \}\).

Then, \(y_k \leq M_n\) for every \(k \geq n\).

Then, \(x_k \leq y_k \leq M_n\) for every \(k \geq n\).

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

Thus, \(m_n \leq M_n\) for every \(n\).

Thus, by Theorem 2.12, \(\lim m_n \leq \lim M_n\).

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#

(Damping sequences)

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

Assume that the following holds:

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

Proof. We proceed as follows.

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

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

Let \(q = u + \epsilon\). Then, \(0 < q < 1\).

By Theorem 2.24, there exists \(n_0 \in \Nat\) such that for all \(n \geq n_0\)

\[ \frac{x_{n+1}}{x_n} < u + \epsilon = q. \]Thus, \(x_{n+1} < q x_n\) for every \(n \geq n_0\).

Let \(x = x_{n_0}\).

Then, we have \(x_n < q^{n - n_0} x\) for every \(n > n_0\).

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

We have \(0 < x_n < q^{n - n_0} x\) for all \(n > n_0\).

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

(Growing sequences)

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

Assume that the following holds:

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

Proof. We proceed as follows.

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

Let \(q = l - \epsilon\). Then, \(q > 1\).

By Theorem 2.25, there exists \(n_0 \in \Nat\) such that for all \(n \geq n_0\)

\[ \frac{x_{n+1}}{x_n} > l - \epsilon = q. \]Thus, \(x_{n+1} > q x_n\) for every \(n \geq n_0\).

Let \(x = x_{n_0}\).

Then, we have \(x_n > q^{n - n_0} x\) for every \(n > n_0\).

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

We have \(x_n > q^{n - n_0} x\) for all \(n > n_0\).

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

Thus, \(\lim x_n = \infty\).