sandwich rule

Sandwich rule

Let $(a_n),(b_n),(c_n)\subset \mathbb{R}$ be sequences with

$$a_n\le b_n\le c_n\text{for all }n\in \mathbb{N}$$

If $(a_n)$ and $(c_n)$ converge to the same limit $a$, then $(b_n)$ also converges to $a$:

$$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{c}_n=a⟹\underset{n\to \infty }{\lim }{b}_n=a$$

The condition can be relaxed to $a_n\le b_n\le c_n$ for all $n\ge N$ (finitely many exceptions don’t matter).

Proof

Let $\epsilon >0$. Since $a_n\to a$ and $c_n\to a$, there exists $n_0$ such that for all $n\ge n_0$:

$$a-\epsilon <a_n\le b_n\le c_n<a+\epsilon$$

So $|b_n-a|<\epsilon$ for $n\ge n_0$ (because $|x-a|<\epsilon$ is equivalent to $a-\epsilon <x<a+\epsilon$; ”$x$ is within distance $ϵ$ of $a$”).

${\lim }_{n\to \infty }\sqrt[n]{a^n+b^n}=\max \{a,b\}$ for $a,b\ge 0$

Assume $b\ge a$ (so we want to show the limit is $b$).

Lower bound: $a^n+b^n\ge b^n$, so $\sqrt[n]{a^n+b^n}\ge \sqrt[n]{b^n}=b$.
(Note: $\sqrt[n]{b^n}=b$ is just algebra: $(b^n{)}^{1/n}=b$. No limit involved.)

Upper bound: Since $a\le b$, we have $a^n\le b^n$, so $a^n+b^n\le 2b^n$.
Thus $\sqrt[n]{a^n+b^n}\le \sqrt[n]{2b^n}=\sqrt[n]{2}\cdot b$.

As $n\to \infty$, $\sqrt[n]{2}\to 1$, so the upper bound $\to b$.

$$b=\sqrt[n]{b^n}\le \sqrt[n]{a^n+b^n}\le \sqrt[n]{2b^n}=\sqrt[n]{2}\cdot b=b$$

By sandwich: $\sqrt[n]{a^n+b^n}\to b=\max \{a,b\}$.

The larger base dominates exponentially.

Sequence without closed-form terms

Consider $b_n=\frac{(1+\sin n{)}^n}{n\cdot 2^n}$. There’s no simple pattern or closed form for $(1+\sin n{)}^n$, so we can’t compute the limit directly.

But we can bound it: $\sin (n)\le 1⟹1+\sin (n)\le 2⟹(1+\sin n{)}^n\le 2^n$.

Therefore:

$$0\le b_n=\frac{(1+\sin n{)}^n}{n\cdot 2^n}\le \frac{2^n}{n\cdot 2^n}=\frac{1}{n}$$

Both $0$ and $\frac{1}{n}$ converge to $0$, so $b_n\to 0$ by sandwich.