# Riemann sum limit

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?

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 $-1/2$.
• 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 $1-e^{1/n^2} = -\dfrac{1}{n^2} + o (\dfrac{1}{n^2})$. – Rigel May 17 '17 at 7:02