# Math for Non-Geeks/The squeeze theorem

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

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.
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

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
$a_{n}={\frac {(-1)^{n}}{n^{2}}}$
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).
Math for Non-Geeks: Template:Aufgabe

#### 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.
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 $\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\tfrac {n!}{n^{n}}}=0$.
Math for Non-Geeks: Template:Aufgabe

#### 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 $\lim _{n\to \infty }a_{n}'=3$.
Math for Non-Geeks: Template:Aufgabe

#### 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 $\lim _{n\to \infty }a_{n}={\tfrac {1}{2}}$.
Math for Non-Geeks: Template:Aufgabe

#### Squeeze theorem: example & exercise 5

Math for Non-Geeks: Template:Aufgabe
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:
${\begin{aligned}\lim _{n\to \infty }\left(1-{\tfrac {1}{n^{2}}}\right)^{n}&=\lim _{n\to \infty }\left(\left(1+{\tfrac {1}{n}}\right)\left(1-{\tfrac {1}{n}}\right)\right)^{n}\\&=\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}\left(1-{\tfrac {1}{n}}\right)^{n}\\&=\lim _{n\to \infty }\left(1+{\tfrac {1}{n}}\right)^{n}\cdot \lim _{n\to \infty }\left(1-{\tfrac {1}{n}}\right)^{n}\\&=e\cdot {\tfrac {1}{e}}=1\end{aligned}}$
Math for Non-Geeks: Template:Aufgabe

## 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 $\lim _{n\to \infty }{\sqrt[{n}]{c}}=1$.
Math for Non-Geeks: Template:Aufgabe