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Find, with proof, $\lim\limits_{n\to\infty} \sum_{i=0}^n (1-i/n)^n$.

Intuitively the limit should just be $\sum_{i=0}^\infty e^{-i} = \frac{1}{1-e^{-1}}.$ But one needs to justify why one can take the limit of both terms in the sequence. If one strictly works from the definition, then one needs to prove that for all $\epsilon > 0,$ there exists $N$ so that for $n\ge N, |\sum_{i=0}^n (1-i/n)^n - \frac{1}{1-e^{-1}}| < \epsilon.$ First choose $M$ so that $\sum_{i=n}^\infty e^{-i}=\frac{e^{-n}}{1-e^{-1}} < \epsilon/2$ for $n\ge M$. We have $|\sum_{i=0}^n (1-i/n)^n - \frac{1}{1-e^{-1}}| = |\sum_{i=0}^n (1-i/n)^n - \sum_{i=0}^n e^{-i} - \sum_{i>n} e^{-i}|.$ For each $i$, we can choose $n$ so that $|(1-i/n)^n - e^{-i}| < \epsilon/2^i,$ but that's where I'm stuck; we might have to choose infinitely many $n$'s, whereas we need to find a maximum of finitely many $n$'s.

Fred Jefferson
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1 Answers1

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Here's the basic outlines of a proof: use a well-known identity relating your expression and $e^{-k}$, use these inequalities to bound your sum, conclude by the squeeze theorem what the limit is. I'll skirt over some of the more difficult bookkeeping details but the basic idea is sound.

It is well known that

$$\lim_{n\to\infty} \left(1-\frac{k}{n}\right)^n=e^{-k}$$

and that

$$\left(1-\frac{k}{n}\right)^n<e^{-k}<\left(1-\frac{k}{n}\right)^{n-1}$$

$$\Rightarrow \left(1-\frac{k}{n}\right)e^{-k}<\left(1-\frac{k}{n}\right)^n$$

for large enought $n$. This implies that for these large $n$ we have

$$\sum_{k=1}^n \left(1-\frac{k}{n}\right)e^{-k}<\sum_{k=1}^n\left(1-\frac{k}{n}\right)^{n}<\sum_{k=1}^n e^{-k}$$

Now, some of these sums are geometric limits:

$$\Rightarrow \frac{e^{-n}(e^n-1)}{e-1} -\frac{1}{n}\sum_{k=1}^nke^{-k}<\sum_{k=1}^n\left(1-\frac{k}{n}\right)^{n}<\frac{e^{-n}(e^n-1)}{e-1}$$

and that

$$\lim_{n\to\infty}\frac{e^{-n}(e^n-1)}{e-1}=\frac{1}{e-1}$$

What is the sum

$$\frac{1}{n}\sum_{k=1}^nke^{-k}$$

in the limit as $n$ goes to infinity? Well, this is similar to a geometric series and can also be explicitly computed to be

$$=\frac{e^{-n}(n+e^{n+1}-en-e)}{(e-1)^2n}$$

It is not to hard to show that this is

$$\lim_{n\to\infty}\frac{e^{-n}(n+e^{n+1}-en-e)}{(e-1)^2n}=0$$

With the other limits, we have

$$\lim_{n\to\infty}\sum_{k=1}^n \left(1-\frac{k}{n}\right)e^{-k}\leq \lim_{n\to\infty}\sum_{k=1}^n\left(1-\frac{k}{n}\right)^{n}\leq \lim_{n\to\infty}\sum_{k=1}^n e^{-k}$$

$$\Rightarrow \frac{1}{e-1}+0\leq \lim_{n\to\infty}\sum_{k=1}^n\left(1-\frac{k}{n}\right)^{n}\leq\frac{1}{e-1}$$

We conclude

$$\lim_{n\to\infty}\sum_{k=1}^n\left(1-\frac{k}{n}\right)^{n}=\frac{1}{e-1}$$

QC_QAOA
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