TITLE: Find $\lim\limits_{n \to \infty}\frac1{n^3}\sum\limits_{k=0}^{n-1}k^2e^{-\frac{k}{n}}$
QUESTION [5 upvotes]: Can you help me find 

$$\lim_{n \to \infty}\frac1{n^3}\sum_{k=0}^{n-1}k^2e^{-\frac{k}{n}} \ \ ?$$ 

Using Riemann sums, I found it is equal to $2-\frac3{e}$. 
Is that correct ?
Thank you in advance.

REPLY [10 votes]: This is just
$$
\lim_{n \to \infty} \frac 1{n^3} \sum_{k=0}^{n-1} k^2 e^{-k/n} = \lim_{n \to \infty} \sum_{k=0}^{n-1} \left( \frac {k+1}n - \frac kn \right) \left( \frac kn \right)^2 e^{- \left( \frac kn \right) } = \int_0^1 x^2 e^{-x} \, dx.
$$
Do that and you got your answer (integrate by parts twice). Looks like you already did so I won't detail more.
Hope that helps,