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Can this limit be solved with Riemann sums?:

$$\lim_{n\to\infty}\left( n-\sum_{k=1}^ne^{\frac{k}{n^2}}\right) $$

Tried solving it like this:

$$\lim_{n\to\infty}n\left(1-\frac{1}{n}\sum_{k=1}^ne^{\frac{k}{n^2}}\right)$$

and after integrating the sum I got this should be infinity.

Is it correct like this?

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In this case I think that it's better to compute explicitly the geometric sum $$ s_n :=\sum_{k=1}^n e^{k/n^2} = \sum_{k=1}^n (e^{1/n^2})^k = e^{1/n^2} \frac{1- e^{1/n}}{1-e^{1/n^2}}. $$ It is easily seen that $s_n = n + \dfrac{1}{2} + o(1)$, hence your limit is $-1/2$.

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  • $\begingroup$ This is very neat. Good answer $\endgroup$ – Jam May 17 '17 at 6:51
  • $\begingroup$ Could you explain more? I understand taking it as a geometric sum but why do you write it after as n+1/2+o(1) and what does that o mean? $\endgroup$ – Lola May 17 '17 at 6:51
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    $\begingroup$ It's the asymptotic behaviour of $s_n$, obtained writing $1-e^{1/n} = -\dfrac{1}{n} - \dfrac{1}{2n^2} + o(\dfrac{1}{n^2})$ and $1-e^{1/n^2} = -\dfrac{1}{n^2} + o (\dfrac{1}{n^2})$. $\endgroup$ – Rigel May 17 '17 at 7:02

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