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.
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.
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,