Squeeze (Sandwich, Pinching ...) Theorem (Lemma):
An Example

A very useful theorem that is known under many different names tells us that if two functions have the same limit and a third function has values bounded by the other two, then it also has the same limit:

Assume $g(x)\le f(x)\le h(x)$ on a set that allows passing to the limit as $x\rightarrow a,$ where $a$ could be finite or infinite, and $x$ could be continuous or discrete. Then $\displaystyle\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L$ imply $\displaystyle\lim_{x\rightarrow a}f(x)=L.$

In elementary calculus, the theorem is routinely used to prove $\displaystyle\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$ or $\displaystyle\lim_{x\rightarrow 0}x\sin\frac{1}{x}=0.$ Almost no other examples can be found on the web. But here's one application I have serendipitously come across:

$\displaystyle\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{n\choose k}=2,$

where $\displaystyle {n\choose k}=\frac{n!}{k!(n-k)!}$ is the binomial coefficient,
"$n$ choose $k.$"

Proof

This is common knowledge that in every row the middle term (or two) is a the largest:

Pascal's triangle

Formally, $\displaystyle {n\choose k+1}\ge \displaystyle {n\choose k},$ for $\displaystyle k\le\frac{n-1}{2},$ because of $\displaystyle {n\choose k+1}=\frac{n-k}{k+1}{n\choose k}$ and $\displaystyle\frac{n-k}{k+1}\ge 1$ is equivalent to $\displaystyle k\le\frac{n-1}{2}.$

Thus, for $n\ge 2,$

$\displaystyle 2\le\sum_{k=0}^{n}\frac{1}{n\choose k}\le 2+\frac{2}{{n\choose 1}} + \sum_{k=2}^{n-2}\frac{1}{n\choose 2} = 2+\frac{2}{{n\choose 1}}+\frac{2(n-3)}{n(n-1)},$

and the assertion follows by the Squeeze (Sandwich, Pinching ...) Theorem (Lemma).

Reference

  1. G. Huvent, Sur la sommme $\displaystyle\sum_{k=0}^{n}\frac{1}{n \choose k}$.

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