$$ \lim\limits_{n\to\infty}\frac{\sum\limits_{k=1}^{n}k^n}{n^n}$$
$$ \lim\limits_{n\to\infty}(\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\frac{e}{e-1})\times{n}$$
They comes from internet.
For the first limit, I get the answer from second one, which is $\frac{e}{e-1}$. By observing the format of sum of the geometric series, I find
$\frac{e}{e-1}=\sum\limits_{k=0}^{\infty}e^{-k}$.
So, we need to prove there is a whole number N, for all $n>N$, $|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\sum\limits_{k=0}^{\infty}e^{-k}|<\varepsilon$.
On the one side, $\lim\limits_{n\to\infty}\sum\limits_{k=n}^{\infty}e^{-k}=0 $.
So, we just need to find a whole number N that satisfy for all $n>N$, $|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\sum\limits_{k=0}^{n-1}e^{-k}|<\varepsilon$, because $|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\sum\limits_{k=0}^{\infty}e^{-k}|=|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\sum\limits_{k=0}^{n-1}e^{-k}+\sum\limits_{k=n}^{\infty}e^{-k}|<|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\sum\limits_{k=0}^{\infty}e^{-k}|+|\sum\limits_{k=n}^{\infty}e^{-k}|$.
On the other hand, we can change the format of the sum.
So we have $$|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\sum\limits_{k=0}^{n-1}e^{-k}|=|\frac{\sum\limits_{k=1}^{n}(n+1-k)^n}{n^n}-\sum\limits_{k=0}^{n-1}e^{-k}|=|\frac{\sum\limits_{k=0}^{n-1}(n-k)^n}{n^n}-\sum\limits_{k=0}^{n-1}e^{-k}|=|\sum\limits_{k=0}^{n-1}{(1-\frac{k}{n})^n-e^{-k}}|=|\sum\limits_{k=0}^{n-1}{(1-\frac{k}{n})^n-\lim\limits_{n\to\infty}(1-\frac{k}{n})^n}| $$.
But I am not sure what should I do next. Shall I Use Taylor expansion or something else? I tried Taylor expansion, but I failed.
For the second one, my mind is that we can prove$|\frac{\sum\limits_{k=1}^{n}k^n}{n^n}-\frac{e}{e-1}|=o(n)$, by using Taylor expansion.(I Guess it equals to $0$)
But I can not prove it.
What is more, how can I solve the first problem without knowing the answer?
