$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

The squeeze rule (also known as the Sandwich Rule) is one of the most useful tools in the study of sequences. It allows us to show that a sequence converges to a limit if we ‘sandwich’ it between two other sequences that converge to the same limit. In effect the sequence gets squeezed between the other two.

**Theorem: The Squeeze Rule**

Suppose that $a_n, b_n$ and $c_n$ are sequences such that $a_n\leq b_n \leq c_n$ for all $n$. If $a_n\to l$ and $c_n \to l$, then $b_n \to l$.

**Proof**

Given $\varepsilon >0$ and $a_n\to l$ and $c_n\to l$, there exists $N’$ such that

\[

l-\varepsilon \leq a_n \leq l+\varepsilon {\text{ and }}

l-\varepsilon \leq c_n \leq l+\varepsilon

\]

for all $n>N’$. Thus

\[

l-\varepsilon \leq a_n \leq b_n \leq c_n \leq l+\varepsilon

\]

for all $n>\max \{ N, N’\}$. Hence, $b_n\to l$.

▢

**Theorem**

Let $p\in \N$. Then $n^{-p}\to 0$.

**Proof**

Let $a_n=0$, $b_n=\frac{1}{n^p}$ and $c_n=\frac{1}{n}$.

For $n\geq 1$ we have $n^p\geq n$, and so

\[

0< \frac{1}{n^p}\leq \frac{1}{n}.
\]
Thus, from $1/n\to 0$ and the Squeeze Rule, $n^{-p}\to 0$.

▢

The use of $a_n=0$ as in this proof is quote common.

**Example**

Find the limit of $\frac{\cos(n)}{n}$.

Let $b_n=\frac{\cos(n)}{n} $. Then consider the sequence $\left| b_n\right|$. We have

\[

0\leq \left| b_n\right| = \left| \frac{\cos(n)}{n} \right|

=\left| \cos(n)\right| \left| \frac{1}{n} \right|

\leq \left| \frac{1}{n} \right|

= \frac{1}{n}\to 0.

\]

By the Squeeze Rule $|b_n|\to 0$.

Since $|b_n|\to 0$ we have $b_n\to 0$ too.

**Example**

We have $\lim _{n\to \infty } \sqrt[n]{n} =1$.

We can use the trick of defining a new sequence. Let $b_n=\sqrt[n]{n}-1$.

Then, for $n>1$,

\begin{eqnarray*}

b_n=\sqrt[n]{n}-1 &\iff & (1+b_n)^n = n \\

&\iff & 1 + nb_n + \frac{n(n-1)}{2!} b_n^2 + \dots + b_n^n = n, \\

& & \qquad \qquad \qquad {\text{ by the binomial theorem}},\\

&\Longrightarrow & \frac{n(n-1)}{2!} b_n^2 \leq n \\

&\iff & b_n \leq \sqrt{\frac{2n}{n(n-1)} } , {\text{ as }} b_n\geq 0 ,\\

&\iff & b_n \leq \sqrt{\frac{2}{n-1} } .

\end{eqnarray*}

As $\sqrt[n]{n}\geq 1$ for all $n$, we have $b_n\geq 0$. Therefore,

\[

0\leq b_n \leq \sqrt{\frac{2}{n-1} } .

\]

Since $\frac{2}{n-1} = \frac{2/n}{1- 1/n}$ the Algebra of Limits gives $\sqrt{\frac{2}{n-1}}\to 0$. Hence, by the Squeeze Rule, $b_n\to 0$. Thus $\sqrt[n]{n} \to 1 $ by the Algebra of Limits.