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The problem states as,

Prove the sequence $$S_n= \Bigl(\frac{1}{n}\Bigr)^n+\Bigl(\frac{2}{n}\Bigr)^n+ \cdots \Bigl(\frac{n}{n}\Bigr)^n$$ converges to $\frac{e}{e-1}$.

A simple (possibly wrong) solution:

We can also write the sequence as $$S_n=\Bigl(\frac{n}{n}\Bigr)^n+ \Bigl(\frac{n-1}{n}\Bigr)^n+\Bigl(\frac{n-2}{n}\Bigr)^n+ \cdots +\Bigl(\frac{2}{n}\Bigr)^n+\Bigl(\frac{1}{n}\Bigr)^n$$

Where we just rearranged the terms of the sequence. Since the series $\lim \limits_{n \to \infty} S_n$ is absolutely convergent, the rearrangement will not affect the value of the limit.

Now by passing the limit to $S_n$ $$\lim \limits_{n \to \infty}S_n=\lim \limits_{n \to \infty}\Bigl(\frac{n}{n}\Bigr)^n+ \lim \limits_{n \to \infty}\Bigl(\frac{n-1}{n}\Bigr)^n+\lim \limits_{n \to \infty}\Bigl(\frac{n-2}{n}\Bigr)^n+ \cdots $$ or, $$\lim \limits_{n \to \infty}S_n=1+e^{-1}+e^{-2}+\cdots$$ or, $$\lim \limits_{n \to \infty}S_n= \frac{e}{e-1}$$

Why I'm considering it wrong:

I know the basic limit theorem $\lim(X_1+X_2)=\lim X_1+\lim X_2$ and by mathematical induction, the theorem is true for sum of any finite number of $X_n$'s. However, the theorem does not hold good if $n \to \infty$. For example, let us take a sequence $$a_n= \frac{1}{n^2}+\frac{2}{n^2}+\cdots \frac{n}{n^2}$$

Where $\lim\limits_{n \to \infty}a_n=\frac{1}{2}$, although if we take limit term by term in the summand (i.e expression of $a_n$), we would get $\lim\limits_{n \to \infty}a_n=0$.

Now coming back to our original sequence of concern, it is fairly easy to show that$\lim\limits_{n \to \infty}S_n \leq \frac{e}{e-1}$, since the function $f(x)=\Bigl(1-\frac{k}{x} \Bigr)^x$ is monotonically increasing $\forall x \geq 1$ and $\forall k \in \Bbb N$.

Now we got somehow to show $\lim\limits_{n \to \infty}S_n \geq \frac{e}{e-1}$, in order to use the squeeze theorem to get the desired result. However, I'm unable to make any significant progress to find the same. So my questions are,

i) How to show $\lim\limits_{n \to \infty}S_n \geq \frac{e}{e-1}$ or what can be some alternate method to find the desired limit of the sequence?

ii) Moreover, are there any general results when $\lim(X_1+X_2+ \cdots) = \lim X_1+ \lim X_2+ \cdots$ holds even when the number of terms in the summand are infinity?

L--
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    "Dominated convergence", if $\lvert a_{n}^{(k)}\rvert \leqslant b_n$ for all $n,k$, $\sum b_n < +\infty$, and $\lim_{k \to \infty} a_n^{(k)} = \alpha_n$ for all $n$, then $$\lim_{k \to \infty} \sum_{n = 0}^{\infty} a_n^{(k)} = \sum_{n = 0}^{\infty} \alpha_n$$ (and all series converge absolutely). This applies to $S_n$. – Daniel Fischer May 17 '20 at 15:01
  • I see. I was not familiar with that theorem, thank you for pointing out @DanielFischer. – L-- May 17 '20 at 15:04
  • Yes, your instinct that the argument is wrong is good, but the dominated convergence theorem lets Is make the argument in this case. – Thomas Andrews May 17 '20 at 15:08
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    It is good that you've been sceptical. You've even come up with an example where the analogous argument fails. That's very good. The "simple solution" as it stands is indeed not good, it needs an elaboration why the steps are legitimate. Now, a good exercise is to prove the theorem above, and to check that it indeed applies here. $$a_n^{(k)} = \begin{cases} \bigl(1 - \frac{k}{n}\bigr)^n &\text{if } k < n, \ 0 &\text{if } k \geqslant n.\end{cases}$$ – Daniel Fischer May 17 '20 at 15:10
  • Just a curiosity https://www.wolframalpha.com/input/?i=integral&assumption=%7B%22F%22%2C+%22Integral%22%2C+%22rangestart%22%7D+-%3E%220%22&assumption=%7B%22C%22%2C+%22integral%22%7D+-%3E+%7B%22Calculator%22%7D&assumption=%7B%22F%22%2C+%22Integral%22%2C+%22integrand%22%7D+-%3E%22%28e-1%29W%28x%29%5Ee%2Fx%5Ee%22&assumption=%7B%22F%22%2C+%22Integral%22%2C+%22rangeend%22%7D+-%3E%22infinity%22 . Maybe one way to prove it ? – Miss and Mister cassoulet char May 17 '20 at 15:24
  • I will attempt an intuitive answer - In the first case you actually calculate the impact of the addition of a term. In the second example the conversion is made while adding values where this impact is neglected when you "just" view tern on its own. – Moti May 17 '20 at 16:11

1 Answers1

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I thought it might be instructive to present an approach that does not rely on the Dominated Convergence Theorem. To that end we now proceed.

We begin by fixing a number $N$ with $1\le N<n-1$. Then, we rite $$\sum_{k=0}^{n-1}\left(1-\frac kn\right)^n=\sum_{k=0}^N \left(1-\frac kn\right)^n+\sum_{k=N+1}^{n-1}\left(1-\frac kn\right)^n\tag1$$ Using $\left(1-\frac kn\right)^n<e^{-k}$, we see that the second sum on the right-hand side of $(1)$ is bounded above by $\frac{e^{-N}}{e-1}$. Then, letting $n\to\infty$ we have $$\lim_{n\to \infty}\sum_{k=0}^{n-1}\left(1-\frac kn\right)^n=\frac{e-e^{-N}}{e-1}+O\left(e^{-N}\right)$$ Finally, letting $N\to\infty$ yields

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

as was to be shown!

Mark Viola
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