Help me to solve the limit $\lim_{n \to \infty} \Big[\Big( \frac 1 n\Big)^n+\Big(\frac 2 n\Big)^n + \dots +\Big(\frac n n\Big)^n\Big]=\dots $ Help me with this limit to infinity question (https://i.stack.imgur.com/Jo52z.jpg)
$$\lim_{n \to \infty} \Big[\Big( \frac 1 n\Big)^n+\Big(\frac 2 n\Big)^n + \dots +\Big(\frac n n\Big)^n\Big]=\dots $$
 A: Boy good thing I put in that false answer ! This way I can still solve the problem even after its closes. LOOPHOLE !
Ok so the limit is 
$$x=\lim\limits_{n\to \infty} \sum\limits_{k=0}^n \left(1-\frac{k}{n}\right)^n$$ now if we just take the first $k$ terms as a lower bound be get
$$1+\left(1-\frac{1}{n}\right)^n+\cdots +\left(1-\frac{k}{n}\right)^n\leq x$$
And taking the limit with $k$ fixed gives 
$$1+\frac{1}{e}+\cdots +\frac{1}{e^k}\leq x$$
On the other hand $y_n=\left(1-\frac{k}{n}\right)^n$ is an increasing sequence with $k$ fixed (I checked it with Bernoulli's inequality). So 
$$\left(1-\frac{k}{n}\right)^n\leq \frac{1}{e^k}$$ and thus 
$$\sum\limits_{k=0}^{\infty}\frac{1}{e^k}$$ is an upper bound.
So the limit is 
$$\frac{e}{e-1}$$
A: If $k$ is small compared with $n$,
$(\frac{n-k}{n})^n
=(1-k/n)^n
\approx e^{-k}
$.
So the sum of the
last $k$ terms is about
$\sum_{j=0}^{k-1} e^{-j}
=\dfrac{1-e^{-k}}{1-e^{-1}}
\approx \dfrac1{1-1/e}
$
for large $k$.
Since
$-\ln(1-x) > x$,
$1-x < e^{-x}
$
so
$1-k/n < e^{-k/n}$
or
$(1-k/n)^n < e^{-k}$.
Therefore
the sum of the terms
after the $k$-th is
$\sum_{j=k}^{n}(1-j/n)^n 
\lt \sum_{j=k}^{n-1} e^{-j}
=\dfrac{e^{-k}-e^{-n}}{1-e^{-1}}
\lt \dfrac{e^{-k}}{1-1/e}
\to 0
$
for large $k$.
Therefore,
and I think this can be made
rigorous,
the sum is
$\dfrac1{1-1/e}
\approx  1.5819767068693265$.
Wolfy seems to agree.
