Find $\lim\limits_{n \to \infty} \left ( n - \sum\limits_{k = 1} ^ n e ^{\frac{k}{n^2}} \right)$.

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

Using $e^x=1+x+O(x^2)$ along with $\sum_{k=1}^nk=\frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2=O\left(n^3\right)$ , we assert that

$$\begin{align} \sum_{k=1}^ne^{k/n^2}&=\sum_{k=1}^n\left(1+\frac{k}{n^2}\right)+O\left(\frac{1}{n}\right)\\\\ &=n+\frac{n(n+1)}{2n^2}+O\left(\frac1n\right)\\\\ &=n+\frac12+O\left(\frac1n\right) \end{align}$$

Hence, we see that

$$\lim_{n\to\infty}\left(n-\sum_{k=1}^ne^{k/n^2}\right)=-\frac12$$