Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$. I have to find the limit:
$$\lim\limits_{n \to \infty} \bigg ( n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \bigg)$$
This is what I managed to do:
$$ e^{\frac{1}{n^2}} + e^{\frac{1}{n^2}} + ... + e^{\frac{1}{n^2}}
\le e^{\frac{1}{n^2}} + e^{\frac{2}{n^2}} + ...e^{\frac{n}{n^2}} \le
e^{\frac{n}{n^2}} + e^{\frac{n}{n^2}} + ... e^{\frac{n}{n^2}}$$
$$ n e^{\frac{1}{n^2}}
\le \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
ne^{\frac{1}{n}}$$
$$ -n e^{\frac{1}{n}}
\le - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
- n e^{\frac{1}{n^2}}$$
$$ n - n e^{\frac{1}{n}}
\le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le
n - n e^{\frac{1}{n^2}}$$
Here I found that the limit of the left-hand side is equal to $-1$, while the limit of the right-hand side is $0$. So I got that:
$$-1 \le n - \displaystyle\sum_{k = 1} ^ n e ^{\frac{k}{n^2}} \le 0$$
And I cannot draw a conclusion about the exact limit. What should I do?

Hint: $$a+a^2+...+a^n=\frac{a(a^n-1)}{a-1}$$
