The squeeze theorem

verweis=Math for Non-Geeks

The squeeze theorem is a powerful tool to determine the limit of a complicated sequence. It is based on comparison to simpler sequences, for which the limit is easily determinable.

Motivation

Convergence proof for a root sequence (video in German)

The intuition behind this theorem is quite simple: We are given a complicated sequence $(a_{n})_{n \in ℕ}$ and want to know whether it converges. Often, one can leave out terms in the complicated sequence $(a_{n})_{n \in ℕ}$ and get some simpler sequences $(b_{n})_{n \in ℕ}$ and $(c_{n})_{n \in ℕ}$ . If $(b_{n})_{n \in ℕ}$ is a lower bound and $(c_{n})_{n \in ℕ}$ an upper bound, then $(a_{n})_{n \in ℕ}$ is "caught" in the space between both sequences. If both sequences converge to the same limit $a$ , they "squeeze" together this space to this single point and $(a_{n})_{n \in ℕ}$ has no other option than to converge towards $a$ , as well.

You may also visualize this theorem by a "ham & cheese sandwich" (see image on the right). The upper and lower bounds $(b_{n})_{n \in ℕ}$ and $(c_{n})_{n \in ℕ}$ act as two "slices of toast", which confine $(a_{n})_{n \in ℕ}$ , i.e. the "filling". If you squeeze the toast slices together, the filling in between will also squeeze to this point.

Squeeze theorem (video in German, YouTube video from the YouTube channel Maths CA).

The theorem reads as follows:

Theorem (Squeeze theorem)

Let $(a_{n})_{n \in ℕ}$ be any sequence. If two sequences $(b_{n})_{n \in ℕ}$ and $(c_{n})_{n \in ℕ}$ exist with $b_{n} \leq a_{n} \leq c_{n}$ for all $n \in ℕ$ and $lim_{n \to \infty} b_{n} = lim_{n \to \infty} c_{n} = a$ for some common limit $a \in ℝ$ , then $(a_{n})_{n \in ℕ}$ will also converge towards this limit $a$ .

Proof (Squeeze theorem)

Let $(b_{n})_{n \in ℕ}$ and $(c_{n})_{n \in ℕ}$ be such that $b_{n} \leq a_{n} \leq c_{n}$ and $lim_{n \to \infty} b_{n} = lim_{n \to \infty} c_{n} = a$ for some $a \in ℝ$ .

We need to prove $lim_{n \to \infty} a_{n} = a$ , i.e. for each $ϵ > 0$ there is an $N \in ℕ$ , such that $| a_{n} - a | < ϵ$ for all $n \geq N$ . Let $ϵ > 0$ be arbitrary. By assumption, the convergences $lim_{n \to \infty} b_{n} = a$ and $lim_{n \to \infty} c_{n} = a$ hold.

Hence, there are two thresholds $N_{1} \in ℕ$ and $N_{2} \in ℕ$ with $| b_{n} - a | < ϵ$ for all $n \geq N_{1}$ (lower bounding sequence) and $| c_{n} - a | < ϵ$ for all $n \geq N_{2}$ (upper bounding sequence). Now, there is $b_{n} \leq a_{n} \leq c_{n}$ for $n \in ℕ$ . For each term $a_{n}$ we distinguish the two cases $a_{n} \geq a$ (above the limit) and $a_{n} \leq a$ (below the limit). In case $a_{n} \geq a$ there is

$| \underset{\geq 0}{\underset{⏟}{a_{n} - a}} | = a_{n} - a \overset{a_{n} \leq c_{n}}{\leq} \underset{\geq 0}{\underset{⏟}{c_{n} - a}} = | c_{n} - a |$

The other case $a_{n} \leq a$ is treated analogously:

$| \underset{\leq 0}{\underset{⏟}{a_{n} - a}} | = - (a_{n} - a) = a - a_{n} \overset{a_{n} \geq b_{n}}{\leq} a - b_{n} = - (\underset{\leq 0}{\underset{⏟}{b_{n} - a}}) = | b_{n} - a |$

So if the maximum of both thresholds $N = max {N_{1}, N_{2}}$ is reached or exceeded by $n \geq N$ ( i.e. $n \geq N_{1}$ and $n \geq N_{2}$ hold), then for $a_{n} \geq a$ ,

$| a_{n} - a | \leq | c_{n} - a | < ϵ$

and for $a_{n} \leq a$ ,

$| a_{n} - a | \leq | b_{n} - a | < ϵ$

So in either case $| a_{n} - a | < ϵ$ for all $n \geq N$ , which establishes convergence.

Hint

The inequality $b_{n} \leq a_{n} \leq c_{n}$ does not need to be satisfied for all $n \in ℕ$ . Assume a finite number of sequence elements $a_{n}$ are not "caught between $b_{n}$ and $c_{n}$ ". Then, after some finite $n_{0} \in ℕ$ all of these finitely many elements will have been passed and the further sequence elements $a_{n}$ with $n \geq n_{0}$ will indeed be caught as $b_{n} \leq a_{n} \leq c_{n}$ .

Example

We illustrate the squeeze theorem using the root sequence $(\sqrt[n]{c})_{n \in ℕ}$ , whose limit is $1$ for an arbitrary but fixed constant $c > 0$ .

Two cases need to be distinguished:

Case 1: $c \geq 1$

Let us choose some $n_{0} \in ℕ$ , such that $n_{0} \geq c$ . Then, for all $n \in ℕ$ with $n \geq n_{0}$ there is $c \leq n$ , so $\sqrt[n]{c} \leq \sqrt[n]{n}$ , i.e. we have an upper bound. A lower bound is given by $1 = \sqrt[n]{1} \leq \sqrt[n]{c}$ for any $n \in ℕ$ .

In the previous chapter, we have established the limit $lim_{n \to \infty} 1 = 1 = lim_{n \to \infty} \sqrt[n]{n}$ . So the squeeze theorem yields $lim_{n \to \infty} \sqrt[n]{c} = 1$ for any fixed $c \geq 1$ standing inside the root.

Case 2: $0 < c < 1$

In this case, $\frac{1}{c} \geq 1$ , so we are led to the above case, replacing $c$ by $\frac{1}{c}$ . Hence, also $lim_{n \to \infty} \sqrt[n]{\frac{1}{c}} = 1$ and the limit theorems (quotient) yield:

$lim_{n \to \infty} \sqrt[n]{c} = lim_{n \to \infty} \frac{1}{\sqrt[n]{\frac{1}{c}}} = \frac{1}{lim_{n \to \infty} \sqrt[n]{\frac{1}{c}}} = \frac{1}{1} = 1$

A useful special case

We often encounter 0 as a sequence limit (null sequence). Since the squeeze theorem can be used to prove any $a$ to be a limit of a sequence $(a_{n})_{n \in ℕ}$ , it can also be used for $a = 0$ . In particular, for any $(a_{n})_{n \in ℕ}$ converging to $a$ , the sequence $(| a_{n} - a |)_{n \in ℕ}$ must converge to 0:

Theorem (Special case of the squeeze theorem)

Let $(a_{n})_{n \in ℕ}$ be a real sequence, $a \in ℝ$ , and $(c_{n})_{n \in ℕ}$ a null sequence with $| a_{n} - a | \leq c_{n}$ for all $n \in ℕ$ . Then $lim_{n \to \infty} a_{n} = a$ .

Proof (Special case of the squeeze theorem)

We set ${\tilde{a}}_{n} = | a_{n} - a |$ . For the constant sequence $b_{n} = 0$ there is $b_{n} \leq \tilde{a_{n}} \leq c_{n}$ (absolute values are always non-negative). Since $lim_{n \to \infty} b_{n} = 0 = lim_{n \to \infty} c_{n}$ the squeeze theorem also implies $lim_{n \to \infty} {\tilde{a}}_{n} = 0$ . Therefore, $(| a_{n} - a |)_{n \in ℕ}$ is a null sequence, which is equivalent to the statement that $(a_{n})_{n \in ℕ}$ converges to $a$ .

Hint

This special case of the squeeze theorem is also referred to as direct comparison argument for sequences .

Example

Let us take a look at $lim_{n \to \infty} \sqrt{n^{2} + 1} - n = 0$ . Not a simple case: both $\sqrt{n^{2} + 1}$ and $n$ tend to $\infty$ . But we can show that their difference converges to 0, since it is bounded by: $| \sqrt{n^{2} + 1} - n | \leq \frac{1}{2 n}$

Let us prove this inequality. There is $n^{2} + 1 > n^{2}$ so $\sqrt{n^{2} + 1} > \sqrt{n^{2}} = n$ and the difference is positive, i.e. $\sqrt{n^{2} + 1} - n > 0$ .

Now, we remove the root by squaring it: $\begin{matrix} | \sqrt{n^{2} + 1} - n | \leq \frac{1}{2 n} \\ \Leftrightarrow & \sqrt{n^{2} + 1} \leq n + \frac{1}{2 n} \\ \Leftrightarrow & n^{2} + 1 \leq {(n + \frac{1}{2 n})}^{2} = n^{2} + 1 + \frac{1}{4 n^{2}} \\ \Leftrightarrow & 0 \leq \frac{1}{4 n^{2}} \end{matrix}$

The last inequality holds for all $n \in ℕ$ and implies $| \sqrt{n^{2} + 1} - n | \leq \frac{1}{2 n}$

But now, ${(\frac{1}{2 n})}_{n \in ℕ}$ is a null sequence, so the squeeze theorem proves our assertion.

Squeeze theorem: examples and problems

Squeeze theorem: example & exercise 1

Example (Squeeze theorem 1)

Let us consider the sequence $(a_{n})_{n \in ℕ}$ with

$a_{n} = \frac{(- 1)^{n}}{n^{2}}$

We take a look at the first sequence elements $(- 1, \frac{1}{4}, - \frac{1}{9}, \frac{1}{16}, - \frac{1}{25}, \dots)$ . They seem to be alternately positive and negative, and they tend to zero. We use as lower or upper bound $b_{n} = - \frac{1}{n}$ and $c_{n} = \frac{1}{n}$ . Then, $| a_{n} | = \frac{1}{n^{2}} \leq \frac{1}{n}$ so we caught $b_{n} \leq a_{n} \leq c_{n}$ for all $n \in ℕ$ .

In addition, $lim_{n \to \infty} b_{n} = lim_{n \to \infty} c_{n} = 0$ . The squeeze theorem hence implies $lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{(- 1)^{n}}{n^{2}} = 0$ .

Question: Which further sequences

(b_{n})_{n \in ℕ}

and

(c_{n})_{n \in ℕ}

can be used for catching and squeezing

a_{n}

?

We could also use $b_{n} = - \frac{1}{n^{2}}$ and $c_{n} = \frac{1}{n^{2}}$ . Or $b_{n} = - 1 / n^{s}$ and $c_{n} = 1 / n^{s}$ for any real $0 < s \leq 2$ (not necessarily $1$ or $2$ ).

Exercise (Squeeze theorem 1)

Investigate whether the sequence $({\tilde{a}}_{n})_{n \in ℕ}$ converges, using the squeeze theorem, where

${\tilde{a}}_{n} = \frac{1}{n^{s}}$ with $s \in ℚ^{+}$

Solution (Squeeze theorem 1)

For $s \in ℚ^{+}$ there is a $k \in ℕ$ with $1 / k \leq s \leq k$ . By monotonicity of the power function,

$n^{1 / k} \leq n^{s} \leq n^{k}$

Hence

$\underset{= {\tilde{b}}_{n}}{\underset{⏟}{\frac{1}{n^{k}}}} \leq \underset{= {\tilde{a}}_{n}}{\underset{⏟}{\frac{1}{n^{s}}}} \leq \underset{= {\tilde{c}}_{n}}{\underset{⏟}{\frac{1}{n^{1 / k}}}}$

The limit theorems yield

$lim_{n \to \infty} {\tilde{b}}_{n} = lim_{n \to \infty} \frac{1}{n^{k}} = lim_{n \to \infty} {(\frac{1}{n})}^{k} = 0^{k} = 0$

and the root rule gives $lim_{n \to \infty} {\tilde{c}}_{n} = lim_{n \to \infty} \sqrt[k]{1 / n} = 0$ . Hence, by means of the squeeze theorem $lim_{n \to \infty} {\tilde{a}}_{n} = 0$ .

Squeeze theorem: example & exercise 2

Example (Squeeze theorem 2)

Next, we prove that the sequence $(a_{n})_{n \in ℕ}$ with

$a_{n} = \frac{n!}{n^{n}}$

converges to 0.

This sequence is bounded from below as all elements are positive $a_{n} \geq 0 = b_{n}$ for all $n \in ℕ$ . An upper bound can be found by multiplying out $n!$ and $n^{n}$ :

$a_{n} = \frac{n!}{n^{n}} = \frac{\overset{n factors}{\overset{⏞}{1 \cdot 2 \cdot 3 \cdot \dots \cdot n}}}{\underset{n factors}{\underset{⏟}{n \cdot n \cdot n \cdot \dots \cdot n}}} = \frac{1}{n} \cdot \underset{\leq 1 for n \geq 2}{\underset{⏟}{\frac{2 \cdot 3 \cdot \dots \cdot n}{n \cdot n \cdot \dots \cdot n}}} \leq \underset{= c_{n}}{\underset{⏟}{\frac{1}{n}}} for n \geq 2$

Obviously, both bounding sequences converge to zero: $lim_{n \to \infty} b_{n} = lim_{n \to \infty} c_{n} = 0$ . So the squeeze theorem implies $lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{n!}{n^{n}} = 0$ .

Exercise (Squeeze theorem 2)

Use the squeeze theorem to show that $({\tilde{a}}_{n})_{n \in ℕ}$ and $({\hat{a}}_{n})_{n \in ℕ}$ with

${\tilde{a}}_{n} = \frac{3^{n}}{n!}$
${\hat{a}}_{n} = \frac{2 n}{2^{n}}$

both converge to 0.

Solution (Squeeze theorem 2)

Part 1: The trick is multiplying out again. This time, for $3^{n}$ and $n!$ :

$\underset{= {\tilde{b}}_{n}}{\underset{⏟}{0}} \leq {\tilde{a}}_{n} = \frac{3^{n}}{n!} = \frac{\overset{n factors}{\overset{⏞}{3 \cdot 3 \cdot 3 \cdot \dots \cdot 3}}}{\underset{n factors}{\underset{⏟}{1 \cdot 2 \cdot 3 \cdot \dots \cdot n}}} = \frac{3 \cdot 3}{1 \cdot 2} \cdot \underset{\leq 1 for n \geq 3}{\underset{⏟}{\prod_{j = 3}^{n - 1} \frac{3}{j}}} \cdot \frac{3}{n} \leq \frac{9}{2} \cdot \frac{3}{n} = \underset{= {\tilde{c}}_{n}}{\underset{⏟}{\frac{27}{2 n}}} for n \geq 3$

Since $lim_{n \to \infty} {\tilde{b}}_{n} = lim_{n \to \infty} {\tilde{c}}_{n} = 0$ , the squeeze theorem implies $lim_{n \to \infty} {\tilde{a}}_{n} = lim_{n \to \infty} \frac{3^{n}}{n!} = 0$ .

Part 2:

Again, the elements are positive, i.e. ${\hat{a}}_{n} = \frac{2 n}{2^{n}} \geq 0 = {\hat{b}}_{n}$ . Bounding from above can be done using $2^{n} \geq n^{2}$ for all $n \geq 4$ . This inequality can be shown by induction:

Induction basis: $n = 4$ .

$2^{4} = 16 \geq 16 = 4^{2}$

Induction step: $n \to n + 1$ .

$2^{n + 1} = 2 \cdot 2^{n} \overset{IH}{\geq} 2 \cdot n^{2} = n^{2} + n^{2} = n^{2} + n \cdot n = n^{2} + 2 n + \underset{\geq 2}{\underset{⏟}{(n - 2)}} \cdot \underset{\geq 4}{\underset{⏟}{n}} \geq n^{2} + 2 n + 8 \geq n^{2} + 2 n + 1 = (n + 1)^{2}$

So for all $n \geq 4$ ,

${\hat{a}}_{n} = \frac{2 n}{2^{n}} \overset{2^{n} \geq n^{2}}{\leq} \frac{2 n}{n^{2}} = \underset{= {\hat{c}}_{n}}{\underset{⏟}{\frac{2}{n}}}$

Since $lim_{n \to \infty} {\hat{c}}_{n} = 0$ , the squeeze theorem implies $lim_{n \to \infty} {\hat{a}}_{n} = 0$ .

Squeeze theorem: example & exercise 3

Example (Squeeze theorem 3)

Now, let us consider the root sequences $(a_{n})_{n \in ℕ}$ and $({a_{n}}^{'})_{n \in ℕ}$ with

$a_{n} = \sqrt[n]{n^{2} + n}$

and

${a_{n}}^{'} = \sqrt[n]{2^{n} + 3^{n}}$

First sequence: For $(a_{n})$ , monotonicity of the $n$ -th root implies

$a_{n} = \sqrt[n]{n^{2} + n} \overset{n \leq n^{2}}{\leq} \sqrt[n]{n^{2} + n^{2}} = \underset{= c_{n}}{\underset{⏟}{\sqrt[n]{2 n^{2}}}}$

as well as

$a_{n} = \sqrt[n]{n^{2} + n} \overset{n \geq 0}{\geq} \underset{= b_{n}}{\underset{⏟}{\sqrt[n]{n^{2}}}}$

For those two bounding sequences, the limit theorems imply

$lim_{n \to \infty} b_{n} = lim_{n \to \infty} \sqrt[n]{n^{2}} = lim_{n \to \infty} \sqrt[n]{n} \cdot \sqrt[n]{n} = 1 \cdot 1 = 1$

and

$lim_{n \to \infty} c_{n} = lim_{n \to \infty} \sqrt[n]{2 n^{2}} = lim_{n \to \infty} \sqrt[n]{2} \cdot \sqrt[n]{n} \cdot \sqrt[n]{n} = 1 \cdot 1 \cdot 1 = 1$

The squeeze theorem hence implies $lim_{n \to \infty} a_{n} = 1$ .

Second sequence: For $({a_{n}}^{'})$ the root function is monotone, as well

${a_{n}}^{'} = \sqrt[n]{2^{n} + 3^{n}} \overset{2^{n} \leq 3^{n}}{\leq} \sqrt[n]{3^{n} + 3^{n}} = \underset{= {c_{n}}^{'}}{\underset{⏟}{\sqrt[n]{2 \cdot 3^{n}}}}$

and

${a_{n}}^{'} = \sqrt[n]{2^{n} + 3^{n}} \overset{2^{n} \geq 0}{\geq} \underset{= {b_{n}}^{'}}{\underset{⏟}{\sqrt[n]{3^{n}}}}$

We use the limit theorems for the upper and lower bounding sequences

$lim_{n \to \infty} {b_{n}}^{'} = lim_{n \to \infty} \sqrt[n]{3^{n}} = lim_{n \to \infty} 3 = 3$

and

$lim_{n \to \infty} {c_{n}}^{'} = lim_{n \to \infty} \sqrt[n]{2 \cdot 3^{n}} = lim_{n \to \infty} \sqrt[n]{2} \cdot \sqrt[n]{3^{n}} = 1 \cdot 3 = 3$

So the squeeze theorem implies $lim_{n \to \infty} {a_{n}}^{'} = 3$ .

Exercise (Squeeze theorem 3)

Use the squeeze theorem to determine the limits of the sequences $({\hat{a}}_{n})_{n \in ℕ}$ and $({\overline{a}}_{n})_{n \in ℕ}$ with

${\hat{a}}_{n} = \sqrt[n]{n^{2} - n + 1}$
${\overline{a}}_{n} = \sqrt[n]{3^{n} - 2^{n}}$

Solution (Squeeze theorem 3)

Part 1: For $({\hat{a}}_{n})$ the $n$ -th root is monotone, so

$\underset{= {\hat{a}}_{n}}{\underset{⏟}{\sqrt[n]{n^{2} - n + 1}}} \overset{- n + 1 \leq 0}{\leq} \underset{= {\hat{c}}_{n}}{\underset{⏟}{\sqrt[n]{n^{2}}}}$

further, for $n \geq 2$

$n^{2} \geq 2 n ⟺ \frac{n^{2}}{2} \geq n ⟺ - \frac{n^{2}}{2} \leq - n ⟹ - n + 1 \geq - \frac{n^{2}}{2}$

Hence, there is

$\underset{= {\hat{a}}_{n}}{\underset{⏟}{\sqrt[n]{n^{2} - n + 1}}} \overset{- n + 1 \geq - \frac{n^{2}}{2}}{\geq} \sqrt[n]{n^{2} - \frac{n^{2}}{2}} = \underset{= {\hat{b}}_{n}}{\underset{⏟}{\sqrt[n]{\frac{n^{2}}{2}}}}$

The limit theorems for the bounding sequences yield

$lim_{n \to \infty} {\hat{b}}_{n} = lim_{n \to \infty} \sqrt[n]{\frac{n^{2}}{2}} = lim_{n \to \infty} \sqrt[n]{\frac{1}{2}} \cdot {\sqrt[n]{n}}^{2} = 1 \cdot 1^{2} = 1$

and

$lim_{n \to \infty} {\hat{c}}_{n} = lim_{n \to \infty} \sqrt[n]{n^{2}} = lim_{n \to \infty} {\sqrt[n]{n}}^{2} = 1^{2} = 1$

So the squeeze theorem implies $lim_{n \to \infty} {\hat{a}}_{n} = 1$ .

Part 2: For $({\overline{a}}_{n})$ we again have monotonicity of the $n$ -th root

$\underset{= {\overline{a}}_{n}}{\underset{⏟}{\sqrt[n]{3^{n} - 2^{n}}}} \overset{- 2^{n} \leq 0}{\leq} \sqrt[n]{3^{n}} = \underset{= {\overline{c}}_{n}}{\underset{⏟}{3}}$

Further,

$\underset{= {\overline{a}}_{n}}{\underset{⏟}{\sqrt[n]{3^{n} - 2^{n}}}} = \sqrt[n]{3^{n} (1 - {(\frac{2}{3})}^{n})} \overset{- {(\frac{2}{3})}^{n} \geq - \frac{2}{3}}{\geq} \sqrt[n]{3^{n} \cdot (1 - \frac{2}{3})} = \underset{= {\overline{b}}_{n}}{\underset{⏟}{\sqrt[n]{3^{n} \cdot \frac{1}{3}}}}$

The limit theorems for the bounding sequences render

$lim_{n \to \infty} {\overline{b}}_{n} = lim_{n \to \infty} \sqrt[n]{3^{n} \cdot \frac{1}{3}} = lim_{n \to \infty} \sqrt[n]{\frac{1}{3}} \cdot \sqrt[n]{3^{n}} = 1 \cdot 3 = 3$

And by means of the squeeze theorem $lim_{n \to \infty} {\overline{a}}_{n} = 3$ .

Squeeze theorem: example & exercise 4

Example (Squeeze theorem 4)

Let $(a_{n})_{n \in ℕ}$ be the sequence of finite sums

$a_{n} = \sum_{k = 1}^{n} \frac{k}{n^{2} + k}$

Then, for all $k = 1, \dots, n$ we bound:

$\frac{k}{n^{2} + n} \overset{k \leq n}{\leq} \frac{k}{n^{2} + k} \overset{k \geq 0}{\leq} \frac{k}{n^{2}}$

and hence

$\underset{= b_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k}{n^{2} + n}}} \overset{k \leq n}{\leq} \underset{= a_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k}{n^{2} + k}}} \overset{k \geq 0}{\leq} \underset{= c_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k}{n^{2}}}}$

For the lower bound,

$\begin{matrix} b_{n} & = \sum_{k = 1}^{n} \frac{k}{n^{2} + n} \\ = \frac{\sum_{k = 1}^{n} k}{n^{2} + n} \\ ↓ Gaussian summation \sum_{k = 1}^{n} k = \frac{n (n + 1)}{2} \\ = \frac{\frac{n (n + 1)}{2}}{n^{2} + n} \\ = \frac{n (n + 1)}{2 (n^{2} + n)} \\ = \frac{n^{2} + n}{2 n^{2} + 2 n} \\ = \frac{1 + \frac{1}{n}}{2 + \frac{2}{n}} \\ \to \frac{1}{2} with n \to \infty \end{matrix}$

For the upper bound,

$\begin{matrix} c_{n} & = \sum_{k = 1}^{n} \frac{k}{n^{2}} \\ = \frac{\sum_{k = 1}^{n} k}{n^{2}} \\ ↓ Gaussian summation \sum_{k = 1}^{n} k = \frac{n (n + 1)}{2} \\ = \frac{\frac{n (n + 1)}{2}}{n^{2}} \\ = \frac{n (n + 1)}{2 n^{2}} \\ = \frac{n^{2} + n}{2 n^{2}} \\ = \frac{1 + \frac{1}{n}}{2} \\ \to \frac{1}{2} with n \to \infty \end{matrix}$

So the squeeze theorem implies $lim_{n \to \infty} a_{n} = \frac{1}{2}$ .

Exercise (Squeeze theorem 4)

Investigate the limits of the two sequences of finite sums $({\tilde{a}}_{n})_{n \in ℕ}$ and $({\hat{a}}_{n})_{n \in ℕ}$ defined by

${\tilde{a}}_{n} = \sum_{k = 1}^{n} \frac{k^{2}}{n^{3} + k}$
${\hat{a}}_{n} = \sum_{k = 1}^{n} \frac{k^{2}}{n^{4} - 10 k^{2}}$

using the squeeze theorem.

Solution (Squeeze theorem 4)

Part 1: For all $k = 1, \dots, n$ we have:

$\frac{k^{2}}{n^{3} + n} \overset{k \leq n}{\leq} \frac{k^{2}}{n^{3} + k} \overset{k \geq 0}{\leq} \frac{k^{2}}{n^{3}}$

so we have some bounds

$\underset{= {\tilde{b}}_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k^{2}}{n^{3} + n}}} \overset{k \leq n}{\leq} \underset{= {\tilde{a}}_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k^{2}}{n^{3} + k}}} \overset{k \geq 0}{\leq} \underset{= {\tilde{c}}_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k^{2}}{n^{3}}}}$

For the lower bound,

$\begin{matrix} {\tilde{b}}_{n} & = \sum_{k = 1}^{n} \frac{k^{2}}{n^{3} + n} \\ = \frac{\sum_{k = 1}^{n} k^{2}}{n^{3} + n} \\ ↓ sum formula \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ = \frac{\frac{n (n + 1) (2 n + 1)}{6}}{n^{3} + n} \\ = \frac{n (n + 1) (2 n + 1)}{6 (n^{3} + n)} \\ = \frac{(n^{2} + n) (2 n + 1)}{6 n^{3} + 6 n} \\ = \frac{2 n^{3} + 3 n^{2} + n}{6 n^{3} + 6 n} \\ = \frac{2 + \frac{3}{n} + \frac{1}{n^{2}}}{6 + \frac{6}{n^{2}}} \\ \to \frac{2}{6} = \frac{1}{3} with n \to \infty \end{matrix}$

For the upper bound,

$\begin{matrix} {\tilde{c}}_{n} & = \sum_{k = 1}^{n} \frac{k^{2}}{n^{3}} \\ = \frac{\sum_{k = 1}^{n} k^{2}}{n^{3}} \\ ↓ sum formula \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ = \frac{\frac{n (n + 1) (2 n + 1)}{6}}{n^{3}} \\ = \frac{n (n + 1) (2 n + 1)}{6 n^{3}} \\ = \frac{(n^{2} + n) (2 n + 1)}{6 n^{3}} \\ = \frac{2 n^{3} + 3 n^{2} + n}{6 n^{3}} \\ = \frac{2 + \frac{3}{n} + \frac{1}{n^{2}}}{6} \\ \to \frac{2}{6} = \frac{1}{3} with n \to \infty \end{matrix}$

So the squeeze theorem implies $lim_{n \to \infty} {\tilde{a}}_{n} = \frac{1}{3}$ .

Part 2: For all $n \geq 4$ and $k = 1, \dots, n$ there is:

$0 \leq \frac{k^{2}}{n^{4} - 10 k^{2}} \overset{- 10 k^{2} \geq - 10 n^{2}}{\leq} \frac{k^{2}}{n^{4} - 10 n^{2}}$

so we have the bounds

$\underset{= {\hat{b}}_{n}}{\underset{⏟}{0}} \overset{k \geq 0}{\leq} \underset{= {\hat{a}}_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k^{2}}{n^{4} - 10 k^{2}}}} \leq \underset{= {\hat{c}}_{n}}{\underset{⏟}{\sum_{k = 1}^{n} \frac{k^{2}}{n^{4} - 10 n^{2}}}}$

The lower bound is just 0 and for the upper bound,

$\begin{matrix} {\hat{c}}_{n} & = \sum_{k = 1}^{n} \frac{k^{2}}{n^{4} - 10 n^{2}} \\ = \frac{\sum_{k = 1}^{n} k^{2}}{n^{4} - 10 n^{2}} \\ ↓ sum formula \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ = \frac{\frac{n (n + 1) (2 n + 1)}{6}}{n^{4} - 10 n^{2}} \\ = \frac{2 n^{3} + 3 n^{2} + n}{6 n^{4} - 60 n^{2}} \\ = \frac{\frac{2}{n} + \frac{3}{n^{2}} + \frac{1}{n^{3}}}{6 - \frac{60}{n^{2}}} \\ \to 0 with n \to \infty \end{matrix}$

So the squeeze theorem implies $lim_{n \to \infty} {\hat{a}}_{n} = 0$ .

Squeeze theorem: example & exercise 5

Exercise (Squeeze theorem 5)

Does the sequence $a_{n} = {(1 - \frac{1}{n^{2}})}^{n}$ converge? If yes, what is its limit? Provide proofs for your claims.

How to get to the proof? (Squeeze theorem 5)

Perhaps, you have already encountered the sequence $d_{n} = {(1 + \frac{1}{n})}^{n}$ which converges to Euler's number $e$ . This is an upper bound for $a_{n} = {(1 - \frac{1}{n^{2}})}^{n}$ , so with that knowledge, we may assert that it is bounded above. But what is the limit? Maybe, we should start by computing some initial sequence elements:

$\begin{matrix} a_{10} & = 0.90438 \dots \\ a_{100} & = 0.99004 \dots \\ a_{1000} & = 0.99900 \dots \end{matrix}$

It looks as if $(a_{n})_{n \in ℕ}$ converges to $1$ . This gets particularly clear, if we plot the sequence elements in a diagram:

What makes determining the limit hard is the exponent of $n$ . Otherwise, we could use some limit theorems or an epsilon proof and would be done quite fast. Luckily, there is a trick to get rid of the exponent: Bernoulli's inequality:

${(1 - \frac{1}{n^{2}})}^{n} \geq 1 - n \cdot \frac{1}{n^{2}} = 1 - \frac{1}{n}$

The sequence ${(1 - \frac{1}{n})}_{n \in ℕ}$ is hence a lower bound on $a_{n}$ and it clearly converges to $1$ . We hence set $b_{n} = 1 - \frac{1}{n}$ as a lower bound for the squeeze theorem.

How to get an upper bound converging to 1? This is quite easy, since $(a_{n})_{n \in ℕ}$ is always smaller than $1$ : The expression $1 - \frac{1}{n^{2}}$ lies in $[0, 1)$ and therefore, also every positive integer power of it is smaller than $1$ . This includes ${(1 - \frac{1}{n^{2}})}^{n}$ being smaller than $1$ .

So for the squeeze theorem, we have for all $n \in ℕ$ the inequality:

$1 - \frac{1}{n} \leq {(1 - \frac{1}{n^{2}})}^{n} \leq 1$

Both the lower and the upper bounding sequences converge to the identical limit $1$ . So by means of the squeeze theorem, there must also be $lim_{n \to \infty} {(1 - \frac{1}{n^{2}})}^{n} = 1$ .

Proof (Squeeze theorem 5)

Bernoulli's inequality implies

${(1 - \frac{1}{n^{2}})}^{n} \geq 1 - n \cdot \frac{1}{n^{2}} = 1 - \frac{1}{n}$

And there is

$0 < \frac{1}{n^{2}} \leq 1 \Rightarrow 0 \leq 1 - \frac{1}{n^{2}} \leq 1 \Rightarrow {(1 - \frac{1}{n^{2}})}^{n} \leq 1$

So we get the bounds

$1 - \frac{1}{n} \leq {(1 - \frac{1}{n^{2}})}^{n} \leq 1$

Both bounding sequences converge to $lim_{n \to \infty} 1 - \frac{1}{n} = lim_{n \to \infty} 1 = 1$ , so by means of the squeeze theorem, $lim_{n \to \infty} {(1 - \frac{1}{n^{2}})}^{n} = 1$ which was to be shown.

Hint

If you are allowed to use $lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e$ and $lim_{n \to \infty} {(1 - \frac{1}{n})}^{n} = \frac{1}{e}$ , then you can also obtain the result directly, using the third binomial formula:

$\begin{matrix} lim_{n \to \infty} {(1 - \frac{1}{n^{2}})}^{n} & = lim_{n \to \infty} {((1 + \frac{1}{n}) (1 - \frac{1}{n}))}^{n} \\ = lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} {(1 - \frac{1}{n})}^{n} \\ = lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} \cdot lim_{n \to \infty} {(1 - \frac{1}{n})}^{n} \\ = e \cdot \frac{1}{e} = 1 \end{matrix}$

Exercise (Squeeze theorem 5)

Does the sequence $a_{n} = {(1 + \frac{1}{n^{2}})}^{n}$ converge? If yes, what is its limit? Prove your assertions.

How to get to the proof? (Squeeze theorem 5)

Again, Bernoulli's inequality can be used for getting rid of the exponent $n$ . There is

${(1 + \frac{1}{n^{2}})}^{n} \geq 1 + n \cdot \frac{1}{n^{2}} = 1 + \frac{1}{n}$

so we have a bound from below converging to $1$ . But ${(1 + \frac{1}{n^{2}})}^{n} \geq 1$ is already a lower bound converging to $1$ . So Bernoulli's inequality doesn't help us in this form. What we need is an upper bound converging to $1$ . This can be done by a little trick: we put expressions from the numerator in the denominator and apply Bernoulli's inequality afterwards:

$\begin{matrix} {(1 + \frac{1}{n^{2}})}^{n} & = {(\frac{n^{2} + 1}{n^{2}})}^{n} \\ = \frac{1}{{(\frac{n^{2}}{n^{2} + 1})}^{n}} \\ = \frac{1}{{(\frac{n^{2} + 1 - 1}{n^{2} + 1})}^{n}} \\ = \frac{1}{{(1 - \frac{1}{n^{2} + 1})}^{n}} \\ ↓ Bernoulli's inequality \\ \leq \frac{1}{(1 - \frac{n}{n^{2} + 1})} \\ = \frac{1}{(\frac{n^{2} - n + 1}{n^{2} + 1})} \\ = \frac{n^{2} + 1}{n^{2} - n + 1} \end{matrix}$

Hence, we have an upper bound $a_{n} \leq \frac{n^{2} + 1}{n^{2} - n + 1}$ . Dividing the numerator and denominator of the bound by $n^{2}$ gives a quotient of convergent sequences, both with limit $1$ , and the quotient rule gives $lim_{n \to \infty} \frac{n^{2} + 1}{n^{2} - n + 1} = 1$ . Therefore, we have two bounding sequences $b_{n} = 1 \leq {(1 + \frac{1}{n^{2}})}^{n} \leq \frac{n^{2} + 1}{n^{2} - n + 1} = c_{n}$ converging to $1$ and by the squeeze theorem $lim_{n \to \infty} a_{n} = 1$ .

Proof (Squeeze theorem 5)

Let $b_{n} = 1$ and $c_{n} = \frac{n^{2} + 1}{n^{2} - n + 1}$ . There is $b_{n} \leq a_{n} \leq c_{n}$ :

$\begin{matrix} 1 & \leq {(1 + \frac{1}{n^{2}})}^{n} \\ = {(\frac{n^{2} + 1}{n^{2}})}^{n} \\ = \frac{1}{{(\frac{n^{2}}{n^{2} + 1})}^{n}} \\ = \frac{1}{{(\frac{n^{2} + 1 - 1}{n^{2} + 1})}^{n}} \\ = \frac{1}{{(1 - \frac{1}{n^{2} + 1})}^{n}} \\ ↓ Bernoulli's inequality \\ \leq \frac{1}{(1 - \frac{n}{n^{2} + 1})} \\ = \frac{1}{(\frac{n^{2} - n + 1}{n^{2} + 1})} \\ = \frac{n^{2} + 1}{n^{2} - n + 1} \end{matrix}$

The bounding sequences converge to $lim_{n \to \infty} b_{n} = lim_{n \to \infty} 1 = 1$ and (by means of the limit theorem):

$\begin{matrix} lim_{n \to \infty} c_{n} & = lim_{n \to \infty} \frac{n^{2} + 1}{n^{2} - n + 1} \\ = lim_{n \to \infty} \frac{1 + \frac{1}{n^{2}}}{1 - \frac{1}{n} + \frac{1}{n^{2}}} \\ = \frac{1 + 0}{1 - 0 + 0} = 1 \end{matrix}$

So there is $lim_{n \to \infty} b_{n} = lim_{n \to \infty} c_{n} = 1$ and the squeeze theorem implies $lim_{n \to \infty} a_{n} = 1$ .

Alternative proof (Squeeze theorem 5)

If the exponential series $\sum_{k = 0}^{\infty} \frac{1}{k!} = e$ is known, one can also use the binomial theorem for a bound. A lower bound is given by ${(1 + \frac{1}{n^{2}})}^{n} \geq 1$ . For the upper bound, we use the binomial theorem $(a + b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{k} b^{n - k}$ :

$\begin{matrix} {(1 + \frac{1}{n^{2}})}^{n} & = \sum_{k = 0}^{n} (\binom{n}{k}) {(\frac{1}{n^{2}})}^{k} \cdot 1^{n - k} \\ = 1 + \sum_{k = 1}^{n} \frac{n \cdot (n - 1) \cdot \dots \cdot (n - k + 1)}{k!} \cdot \frac{1}{n^{2 k}} \\ = 1 + \sum_{k = 1}^{n} \frac{1}{k!} \cdot \underset{\leq 1}{\underset{⏟}{\frac{n \cdot (n - 1) \cdot \dots \cdot (n - k + 1)}{n \cdot n \cdot \dots \cdot n}}} \cdot \underset{\leq \frac{1}{n}}{\underset{⏟}{\frac{1}{n^{k}}}} \\ \leq 1 + \frac{1}{n} \cdot \sum_{k = 1}^{n} \frac{1}{k!} \\ \leq 1 + \frac{1}{n} \cdot \underset{= e - 1}{\underset{⏟}{\sum_{k = 1}^{\infty} \frac{1}{k!}}} \\ \leq 1 + \frac{e - 1}{n} \end{matrix}$

This yields a suitable upper bound, and we get $1 \leq a_{n} \leq 1 + \frac{e - 1}{n}$ . The squeeze theorem then implies that $(a_{n})_{n \in ℕ}$ converges to $1$ .

Alternative proof (Squeeze theorem 5)

We may use $lim_{n \to \infty} {(1 + \frac{1}{n})}^{n} = e$ for a direct proof. The sequence ${({(1 + \frac{1}{n^{2}})}^{n^{2}})}_{n \in ℕ}$ is a subsequence of ${({(1 + \frac{1}{n})}^{n})}_{n \in ℕ}$ and converges to the same limit:

$lim_{n \to \infty} {(1 + \frac{1}{n^{2}})}^{n^{2}} = e$ Hence, for some $N \in ℕ$ , there is ${(1 + \frac{1}{n^{2}})}^{n^{2}} \leq 2 e$ for all $n \geq N$ . The sequence $a_{n}$ can be recovered from this by taking the $n^{t h}$ root ${(1 + \frac{1}{n^{2}})}^{n} = \sqrt[n]{{(1 + \frac{1}{n^{2}})}^{n^{2}}}$ . So for all $n \geq N$ :

$\begin{matrix} \begin{matrix} 1 \leq & {(1 + \frac{1}{n^{2}})}^{n^{2}} & \leq 2 e \\ ⟹ & \sqrt[n]{1} \leq & \sqrt[n]{{(1 + \frac{1}{n^{2}})}^{n^{2}}} & \leq \sqrt[n]{2 e} \\ ⟹ & 1 \leq & {(1 + \frac{1}{n^{2}})}^{n} & \leq \sqrt[n]{2 e} \end{matrix} \end{matrix}$

With $lim_{n \to \infty} \sqrt[n]{1} = 1 = lim_{n \to \infty} \sqrt[n]{2 e}$ , the squeeze theorem again implies that $(a_{n})_{n \in ℕ}$ converges to $1$ .

Examples & exercises: squeeze theorem for null sequences

Example (Squeeze theorem for null sequences 1)

First, an easy example: we prove $lim_{n \to \infty} \frac{n - 1}{n + 1} = 1$ .

We estimate $| \underset{= a_{n}}{\underset{⏟}{\frac{n - 1}{n + 1}}} - \underset{= a}{\underset{⏟}{1}} |$ by the null sequence $c_{n} = \frac{2}{n}$ as follows:

$| \frac{n - 1}{n + 1} - 1 | = | \frac{n - 1 - (n + 1)}{n + 1} | = | \frac{- 2}{n + 1} | = \frac{2}{n + 1} \leq \frac{2}{n}$

Since $lim_{n \to \infty} c_{n} = lim_{n \to \infty} \frac{2}{n} = 0$ , $(c_{n})_{n \in ℕ}$ is a null sequence. We can use $1 + c_{n}$ as an upper and $1 - c_{n}$ as a lower bound for $a_{n}$ . The squeeze theorem then implies $lim_{n \to \infty} a_{n} = lim_{n \to \infty} \frac{n - 1}{n + 1} = 1$ .

Example (Squeeze theorem for null sequences 2)

Now, it gets a bit more complicated: We fix any $c > 1$ and prove that $lim_{n \to \infty} \sqrt[n]{c} = 1$ . This can be done by bounding $| \sqrt[n]{c} - 1 | = \sqrt[n]{c} - 1$ from above using a null sequence $(c_{n})$ . The trick is to use Bernoulli's inequality, which will yield the upper bound $c_{n} = \frac{c - 1}{n}$ :

$\begin{matrix} c & = (\sqrt[n]{c})^{n} \\ = (1 + (\sqrt[n]{c} - 1))^{n} \\ ↓ Bernoulli's inequality \\ \geq 1 + n \cdot (\sqrt[n]{c} - 1) \end{matrix}$

This implies

$\begin{matrix} c \geq 1 + n \cdot (\sqrt[n]{c} - 1) & \Leftrightarrow c - 1 \geq n \cdot (\sqrt[n]{c} - 1) \\ \Leftrightarrow \frac{c - 1}{n} \geq \sqrt[n]{c} - 1 \\ \Leftrightarrow \sqrt[n]{c} - 1 \leq \frac{c - 1}{n} \end{matrix}$

So we have

$| \sqrt[n]{c} - 1 | = \sqrt[n]{c} - 1 \leq \frac{c - 1}{n} = (c - 1) \cdot \frac{1}{n} \overset{n \to \infty}{\to} 0$

The squeeze theorem for null sequences directly implies $lim_{n \to \infty} \sqrt[n]{c} = 1$ .

Exercise (Squeeze theorem for null sequences)

Prove that $lim_{n \to \infty} \sqrt[n]{n} = 1$ .

Solution (Squeeze theorem for null sequences)

In order to use the squeeze theorem, we need to bound $| \sqrt[n]{n} - 1 |$ from above by a null sequence. The binomial theorem can be used for this bounding. In case $n \geq 2$ :

$\begin{matrix} n & = (\sqrt[n]{n})^{n} \\ = (1 + (\sqrt[n]{n} - 1))^{n} \\ ↓ binomial theorem \\ = \sum_{k = 0}^{n} (\binom{n}{k}) (\sqrt[n]{n} - 1)^{k} \cdot \underset{= 1}{\underset{⏟}{1^{n - k}}} \\ = \underset{= 1}{\underset{⏟}{(\binom{n}{0}) (\sqrt[n]{n} - 1)^{0}}} + \underset{\geq 0}{\underset{⏟}{(\binom{n}{1}) (\sqrt[n]{n} - 1)^{1}}} + (\binom{n}{2}) (\sqrt[n]{n} - 1)^{2} + \underset{\geq 0}{\underset{⏟}{\sum_{k = 3}^{n} (\binom{n}{k}) (\sqrt[n]{n} - 1)^{k}}} \\ \geq 1 + (\binom{n}{2}) (\sqrt[n]{n} - 1)^{2} \\ = 1 + \frac{n (n - 1)}{2} (\sqrt[n]{n} - 1)^{2} \end{matrix}$

This inequality implies

$\begin{matrix} n \geq 1 + \frac{n (n - 1)}{2} \cdot (\sqrt[n]{n} - 1)^{2} & ⟺ n - 1 \geq \frac{n (n - 1)}{2} \cdot (\sqrt[n]{n} - 1)^{2} \\ ⟺ \frac{2}{n} \geq (\sqrt[n]{n} - 1)^{2} \\ ⟺ | \sqrt[n]{n} - 1 | \leq \sqrt{\frac{2}{n}} \end{matrix}$

So the desired bound reads

$| \sqrt[n]{n} - 1 | \leq \sqrt{2} \cdot \frac{1}{\sqrt{n}} \overset{n \to \infty}{\to} 0$

And by means of the squeeze theorem $lim_{n \to \infty} \sqrt[n]{n} = 1$ , which was to be shown.

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