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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.

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uhm - since you know the answer -- what do you want to know, then? – user20266 Dec 31 '11 at 16:51
Just to make sure my answer is right :-) Wolfram Alpha doesn't seem to accept the command. – user20010 Dec 31 '11 at 16:55
The devil is often in the details, so it's possible for your work to be subtly incorrect even if you got the right answer. It would be more beneficial to you, and quicker for the answerers, if you include the entire work rather than just the final answer. – Srivatsan Dec 31 '11 at 17:00
up vote 10 down vote accepted

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,

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Actually, the correct answer is $2-\frac{5}{e}$ and Srivatasan was right ! – user20010 Dec 31 '11 at 17:11
Well I didn't even bother computing the integral... it looks quite elementary. If OP wanted the answer he could've just computed this integral in Wolfram Alpha if he didn't trust himself. =) – Patrick Da Silva Dec 31 '11 at 17:48

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