Regarding evaluation of the limit of the sequence $\Bigl(\frac{1}{n}\Bigr)^n+\Bigl(\frac{2}{n}\Bigr)^n+ \cdots \Bigl(\frac{n}{n}\Bigr)^n$ [duplicate]

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?

Solution 1:

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!