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

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.

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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!

We put, $$ f_n(x)=\left(1-\frac{\lfloor x\rfloor}{n}\right)^n\chi_{[0,n+1]}(x) $$ We have, $$ \int\limits_{[0,+\infty)} f_n(x)d\mu(x)=\sum\limits_{k=0}^n\int\limits_{[k,k+1)}\left(1-\frac{\lfloor x\rfloor}{n}\right)^nd\mu(x)= \sum\limits_{k=0}^n\left(1-\frac{k}{n}\right)^n $$ $$ \lim\limits_{n\to\infty}f_n(x)=\lim\limits_{n\to\infty}\left(1-\frac{\lfloor x\rfloor}{n}\right)^n\cdot \lim\limits_{n\to\infty}\chi_{[0,n+1]}(x)=e^{\lfloor x\rfloor} $$ $\{f_n:n\in\mathbb{N}\}$ is a non-decreasing sequence of non-negative functions, then we get by monotone convergence theorem : $$ \lim\limits_{n\to\infty}\sum\limits_{k=0}^n\left(\frac{k}{n}\right)^n= \lim\limits_{n\to\infty}\sum\limits_{k=0}^n\left(1-\frac{k}{n}\right)^n= \lim\limits_{n\to\infty}\int\limits_{[0,+\infty)} f_n(x)d\mu(x)= $$ $$ \int\limits_{[0,+\infty)} \lim\limits_{n\to\infty}f_n(x)d\mu(x)= \int\limits_{[0,+\infty)} e^{\lfloor x\rfloor}d\mu(x)= \sum\limits_{k=0}^\infty e^{-k}=\frac{1}{1-e^{-1}} $$