Does this limit exist and what's its value? $\lim_{n\to\infty}n^y\sum_ {i=1}^{n}\left[e^{-i}-\left(1-\frac{i}{n}\right)^{\!n}\right]$ Find $$\lim_{n\to \infty}\left\{n^y \sum_ {i=1}^{n}\left[\;e^{-i} -\left(1- \frac {i} {n}\right)^{\!n}\;\right]\right\}$$
I became interested in this problem because the YouTube series BlackPenRedPen (as part of the solution of another limit problem: https://www.youtube.com/watch?v=nPNB26hxLPc&t=7s) solved the problem in the case $y = 0$ by interchanging sum and limit without justification. I realized that the methods used to correctly solve the case $y = 0$ (i.e., the Monotone Convergence Theorem) might solve the stated problem. Before starting to work on the the problem, I was trying to ascertain if the solution was known or easier than I thought.
This problem has been criticized because I didn't supply the reason I was interested in its solution. If one needs a reason to study mathematical questions (beyond interest), then we we should stop working on The Twin Prime Conjecture, Goldbach's Conjecture, the Collatz Problem, the ABC Conjecture, etc.
 A: Evaluation of the Limit
To Approximate
$$
\sum_{k=1}^n\left[\,e^{-k}-\left(1-\frac kn\right)^n\,\right]\tag1
$$
note that
$$
n\log\left(1-\frac kn\right)=-k-\frac{k^2}{2n}-\frac{k^3}{3n^2}-\frac{k^4}{4n^3}-\cdots\tag2
$$
Immediately, $(2)$ implies that $\left(1-\frac kn\right)^n\lt e^{-k}$, so each term in $(1)$ is positive, and therefore, less than $e^{-k}$. Thus,
$$
\begin{align}
\sum_{k=1}^n\left[\,e^{-k}-\left(1-\frac kn\right)^n\,\right]
&=\sum_{k=1}^{n^{1/3}}\left[\,e^{-k}-\left(1-\frac kn\right)^n\,\right]+O\!\left(e^{-n^{1/3}}\right)\\
&=\sum_{k=1}^{n^{1/3}}e^{-k}\left[\,1-e^{-\frac{k^2}{2n^{\vphantom{1}}}-\frac{k^3}{3n^2}-\frac{k^4}{4n^3}-\cdots}\,\right]+O\!\left(e^{-n^{1/3}}\right)\\
&=\sum_{k=1}^{n^{1/3}}e^{-k}\left[\,\frac{k^2}{2n}+O\!\left(\frac{k^4}{n^2}\right)\,\right]\\
&=\frac1{2n}\sum_{k=1}^\infty k^2e^{-k}+O\!\left(\frac1{n^2}\right)\\
&=\frac1{2n}\frac{e^2+e}{(e-1)^3}+O\!\left(\frac1{n^2}\right)\tag3
\end{align}
$$
Therefore,
$$
\bbox[5px,border:2px solid #C0A000]{n\sum_{k=1}^n\left[\,e^{-k}-\left(1-\frac kn\right)^n\,\right]
=\frac{e^2+e}{2(e-1)^3}+O\!\left(\frac1n\right)}\tag4
$$

Calculation of the Sum Used Above
$$
\begin{align}
\sum_{k=1}^\infty k^2e^{-k}
&=\sum_{k=1}^\infty\left[2\binom{k}{2}+\binom{k}{1}\right]e^{-k}\\
&=\sum_{k=1}^\infty\left[2\binom{k}{k-2}+\binom{k}{k-1}\right]e^{-k}\\
&=\sum_{k=1}^\infty\left[2(-1)^{k-2}\binom{-3}{k-2}+(-1)^{k-1}\binom{-2}{k-1}\right]e^{-k}\\
&=\sum_{k=0}^\infty2(-1)^k\binom{-3}{k}e^{-k-2}+\sum_{k=0}^\infty(-1)^k\binom{-2}{k}e^{-k-1}\\
&=\frac{2e^{-2}}{(1-1/e)^3}+\frac{e^{-1}}{(1-1/e)^2}\\
&=\frac{e^2+e}{(e-1)^3}\tag5
\end{align}
$$
A: To determine the behavior of the limit, it suffices to identify the single value of $y$ for which the limit converges to a positive real number, if exists. We claim that $y = 1$ works. More precisely, let
$$ a^{(n)}_k = n \left( e^{-k} - \left( 1 - \frac{k}{n} \right)^n \right). $$
Then we claim that
$$ \lim_{n\to\infty} \sum_{k=1}^{n} a^{(n)}_k = \frac{e(e+1)}{2(e-1)^3} \approx 0.996147. \tag{1} $$
To this end, the following lemma will be quite useful.

Lemma. There exists a constant $ c> 0$ such that
$$ 0 \leq 1 - e^x \left( 1 - \frac{x}{n} \right)^n \leq \frac{cx^2}{n} $$
holds for all $n \geq 1$ and $0 \leq x \leq n$. Moreover,
$$ \lim_{n\to\infty} n \left( 1 - e^x \left( 1 - \frac{x}{n} \right)^n \right) = \frac{x^2}{2}. $$

Accepting the lemma for a moment, we find that $ 0 \leq a^{(n)}_k \leq c k^2 e^{-k} $. Since $\sum_{k=1}^{\infty} ck^2 e^{-k} $ converges, either by the Weierstrass M-test or by the Dominated Convergence Theorem, we get
\begin{align*}
\lim_{n\to\infty} \sum_{k=1}^{n} a^{(n)}_k
= \sum_{k=1}^{\infty} \lim_{n\to\infty} a^{(n)}_k
= \sum_{k=1}^{\infty} \frac{1}{2} k^2 e^{-k}
= \frac{e(e+1)}{2(e-1)^3}.
\end{align*}
So it remains to prove the lemma:
Proof of Lemma. Fix $n \geq 1$ and let
$$ f(x) = 1 - e^x \left( 1 - \frac{x}{n} \right)^n. $$
Then by using the inequality $1+t\leq e^t$ which holds for any $t\in\mathbb{R}$, for $0 \leq x \leq n$ we get
$$ f'(x)
= e^x \left(1 - \frac{x}{n}\right)^{n-1} \frac{x}{n}
\leq e^{x-(n-1)\frac{x}{n}} \frac{x}{n}
\leq \frac{ex}{n}. $$
It is also clear that $f'(x) \geq 0$ on this range of $x$. Therefore the inequality follows with $c = e/2$.  As for the limit, for each fixed $x \geq 0$ we get
\begin{align*}
1 - e^x \left( 1 - \frac{x}{n} \right)^n
&= 1 - \exp\left( x + n \log\left(1 - \frac{x}{n} \right) \right) \\
&= 1 - \exp\left( - \frac{x^2}{2n} + \mathcal{O}\left(\frac{1}{n^2}\right) \right) \\
&= \frac{x^2}{2n} + \mathcal{O}\left(\frac{1}{n^2}\right)
\end{align*}
as $n\to\infty$, and so, the desired conclusion follows.
