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

image about the principle of the sandwich lemma

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

The squeeze theorem

Squueze theorem (video in German, YouTube videofrom the YouTube channel Maths CA).

The theorem reads as follows:

Theorem (Squeeze theorem)

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

Proof (Squeeze theorem)

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

We need to prove $\lim _{n\to \infty }a_{n}=a$ , i.e. for each $\epsilon >0$ there is an $N\in \mathbb {N}$, such that $|a_{n}-a|<\epsilon$ for all $n\geq N$ . Let $\epsilon >0$ 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 $N_{1}\in \mathbb {N}$ and $N_{2}\in \mathbb {N}$ with $|b_{n}-a|<\epsilon$ for all $n\geq N_{1}$ (lower bounding sequence) and $|c_{n}-a|<\epsilon$ for all $n\geq N_{2}$ (upper bounding sequence). Now, there is $b_{n}\leq a_{n}\leq c_{n}$ for $n\in \mathbb {N}$. For each $(a_{n})_{n\in \mathbb {N} }$ 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

$|\underbrace {a_{n}-a} _{\geq 0}|=a_{n}-a{\overset {a_{n}\leq c_{n}}{\leq }}\underbrace {c_{n}-a} _{\geq 0}=|c_{n}-a|$

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

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

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

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

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

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

So in either case $|a_{n}-a|<\epsilon$ 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 $n\in \mathbb {N}$ . Assume, a finite amount of sequence elements $a_{n}$ is not "caught between $b_{n}$ and $c_{n}$". Then, after some finite $n_{0}\in \mathbb {N}$ all of these finite elements will have been passed and the further sequence elements $n\geq n_{0}$ will indeed be caught as $b_{n}\leq a_{n}\leq c_{n}$.

Example

We illustrate the squeeze theorem using the root sequence $\lim _{n\to \infty }{\sqrt[{n}]{c}}=1$ for an arbitrary but fixed constant $c>0$.

Two cases need to be distinguished:

Case 1: $c\geq 1$

Let us choose some $n_{0}\in \mathbb {N}$, such that $n_{0}\geq c$. Then, for all $n\in \mathbb {N}$ 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 $n\in \mathbb {N}$.

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 $c$ standing inside the root.

Case 2: $c<1$

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

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

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 $(a_{n})_{n\in \mathbb {N} }$, it can also be used for $a=0$. Especially, for any convergent $(a_{n})_{n\in \mathbb {N} }$, the sequence $(|a_{n}-a|)_{n\in \mathbb {N} }$ must converge to 0:

Theorem (Special case of the squeeze theorem)

Let $(c_{n})_{n\in \mathbb {N} }$ be a null sequence and $|a_{n}-a|\leq c_{n}$ for all $n\in \mathbb {N}$. 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 positive). 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, $(|a_{n}-a|)_{n\in \mathbb {N} }$ is a null sequence, which is equivalent to the statement that $(a_{n})_{n\in \mathbb {N} }$ 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 $n$ converge to $\infty$ . But we can show that its difference converges to 0, since it is bounded by: $|{\sqrt {n^{2}+1}}-n|\leq {\frac {1}{2n}}$

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: ${\begin{aligned}&\ \ |{\sqrt {n^{2}+1}}-n|\leq {\frac {1}{2n}}\\[0.3em]\Leftrightarrow &\ \ {\sqrt {n^{2}+1}}\leq n+{\frac {1}{2n}}\\[0.3em]\Leftrightarrow &\ \ n^{2}+1\leq \left(n+{\frac {1}{2n}}\right)^{2}=n^{2}+1+{\frac {1}{4n^{2}}}\\[0.3em]\Leftrightarrow &\ \ 0\leq {\frac {1}{4n^{2}}}\end{aligned}}$

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

But now, $\left({\tfrac {1}{2n}}\right)_{n\in \mathbb {N} }$ 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 $(a_{n})_{n\in \mathbb {N} }$ with

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

The first 6 elements of the sequence ${\color {Blue}\left({\tfrac {(-1)^{n}}{n^{2}}}\right)_{n\in \mathbb {N} }}$, ${\color {red}\left(-{\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}$ and ${\color {green}\left({\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}$

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

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

Question: Which further sequences $(b_{n})_{n\in \mathbb {N} }$ and $(c_{n})_{n\in \mathbb {N} }$ can be used for catching and squeezing $a_{n}$?

We could also use $b_{n}=-{\tfrac {1}{n^{2}}}$ and $c_{n}={\tfrac {1}{n^{2}}}$. Or any other power between 0 and 2 (not necessarily 1 or 2).

Squeeze theorem: example & exercise 2

Example (Squeeze theorem 2)

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

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

converges to 0.

The first elements of ${\color {Blue}\left({\tfrac {n!}{n^{n}}}\right)_{n\in \mathbb {N} }}$, ${\color {red}\left(0\right)_{n\in \mathbb {N} }}$ and ${\color {green}\left({\tfrac {1}{n}}\right)_{n\in \mathbb {N} }}$

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

$a_{n}={\frac {n!}{n^{n}}}={\frac {\overbrace {1\cdot 2\cdot 3\cdot \ldots \cdot n} ^{n{\text{ factors}}}}{\underbrace {n\cdot n\cdot n\cdot \ldots \cdot n} _{n{\text{ factors}}}}}={\frac {1}{n}}\cdot \underbrace {\frac {2\cdot 3\cdot \ldots \cdot n}{n\cdot n\cdot \ldots \cdot n}} _{\leq 1{\text{ for }}n\geq 2}\leq \underbrace {\frac {1}{n}} _{=c_{n}}{\text{ für }}n\geq 2$

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 $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {n!}{n^{n}}}=0$.

Squeeze theorem: example & exercise 3

Example (Squeeze theorem 3)

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

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

as well as

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

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

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

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

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

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

and

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

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 $\lim _{n\to \infty }a_{n}'=3$.

Squeeze theorem: example & exercise 4

Example (Squeeze theorem 4)

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

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

Then, for all $k=1,\ldots ,n$ 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

$\underbrace {\sum _{k=1}^{n}{\frac {k}{n^{2}+n}}} _{=b_{n}}{\overset {k\leq n}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k}{n^{2}+k}}} _{=a_{n}}{\overset {k\geq 0}{\leq }}\underbrace {\sum _{k=1}^{n}{\frac {k}{n^{2}}}} _{=c_{n}}$

For the lower bound,

${\begin{aligned}b_{n}&=\sum _{k=1}^{n}{\frac {k}{n^{2}+n}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k}{n^{2}+n}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{Gaussian summation }}\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)}{2}}{n^{2}+n}}\\[0.5em]&={\frac {n(n+1)}{2(n^{2}+n)}}\\[0.5em]&={\frac {n^{2}+n}{2n^{2}+2n}}\\[0.5em]&={\frac {1+{\frac {1}{n}}}{2+{\frac {2}{n}}}}\\[0.5em]&\to {\frac {1}{2}}{\text{ with }}n\to \infty \end{aligned}}$

For the upper bound,

${\begin{aligned}c_{n}&=\sum _{k=1}^{n}{\frac {k}{n^{2}}}\\[0.5em]&={\frac {\sum _{k=1}^{n}k}{n^{2}}}\\[0.5em]&\ {\color {OliveGreen}\left\downarrow \ {\text{Gaussian summation }}\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}\right.}\\[0.5em]&={\frac {\frac {n(n+1)}{2}}{n^{2}}}\\[0.5em]&={\frac {n(n+1)}{2n^{2}}}\\[0.5em]&={\frac {n^{2}+n}{2n^{2}}}\\[0.5em]&={\frac {1+{\frac {1}{n}}}{2}}\\[0.5em]&\to {\frac {1}{2}}{\text{ with }}n\to \infty \end{aligned}}$

So the squeeze theorem implies $\lim _{n\to \infty }a_{n}={\tfrac {1}{2}}$.

Squeeze theorem: example & exercise 5

Hint

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

${\begin{aligned}\lim _{n\to \infty }\left(1-{\tfrac {1}{n^{2}}}\right)^{n}&=\lim _{n\to \infty }\left(\left(1+{\tfrac {1}{n}}\right)\left(1-{\tfrac {1}{n}}\right)\right)^{n}\\&=\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}\left(1-{\tfrac {1}{n}}\right)^{n}\\&=\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}\cdot \lim _{n\to \infty }\left(1-{\tfrac {1}{n}}\right)^{n}\\&=e\cdot {\tfrac {1}{e}}=1\end{aligned}}$

Examples & exercises: squeeze theorem for null sequences

Example (Squeeze theorem for null sequences 1)

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

We estimate $|\underbrace {\tfrac {n-1}{n+1}} _{=a_{n}}-\underbrace {1} _{=a}|$ by the null sequence $c_{n}={\tfrac {2}{n}}$ as follows:

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

Since $\lim _{n\to \infty }c_{n}=\lim _{n\to \infty }{\tfrac {2}{n}}=0$ is a null sequence, we can use $c_{n}$ as an upper and $-c_{n}$ as a lower bound for $a_{n}$. The squeeze theorem then implies $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {n-1}{n+1}}=1$.

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 us to the upper bound $c_{n}={\tfrac {c-1}{n}}$ :

${\begin{aligned}c&=({\sqrt[{n}]{c}})^{n}\\[0.5em]&=(1+({\sqrt[{n}]{c}}-1))^{n}\\[0.5em]&{\color {OliveGreen}\left\downarrow {\text{ Bernoulli's inequality}}\right.}\\[0.5em]&\geq 1+n\cdot ({\sqrt[{n}]{c}}-1)\end{aligned}}$

Which implies

${\begin{aligned}c\geq 1+n\cdot ({\sqrt[{n}]{c}}-1)&\Leftrightarrow c-1\geq n\cdot ({\sqrt[{n}]{c}}-1)\\[0.5em]&\Leftrightarrow {\tfrac {c-1}{n}}\geq {\sqrt[{n}]{c}}-1\\[0.5em]&\Leftrightarrow {\sqrt[{n}]{c}}-1\leq {\tfrac {c-1}{n}}\end{aligned}}$

So we have

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

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