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

Question

I stumbled into a mistake when I evaluated $$L=\lim_{n\to\infty} \left(n-\sum_{k=1}^n e^{\frac{k}{n^2}}\right)$$ My first try: $$L=-\lim_{n\to\infty} \sum_{k=1}^{n}\left(e^{\frac{k}{n^2}}-1\right)=-\lim_{n\to\infty} \sum_{k=1}^{n}\left(\frac{e^{\frac{k}{n^2}}-1}{\frac{k}{n^2}}\frac{k}{n^2}\right)=-\lim_{n\to\infty} \sum_{k=1}^{n}\frac{k}{n^2}$$$$L=-\frac{1}{2}\lim_{n\to\infty}\frac{n(n+1)}{n^2}=-\frac12$$ Now this looks pretty clear, however I don't understand what is wrong with the following approach. $$L=\lim_{n\to\infty}(n-(e^\frac{1}{n^2}+e^\frac{2}{n^2}+\cdots + e^\frac{n}{n^2}))=\lim_{n\to\infty} ((1-e^\frac{1}{n^2})+(1-e^\frac{2}{n^2})+\cdots + (1-e^\frac{n}{n^2}))$$ since $$\lim_{n\to\infty} e^\frac{k}{n^2}\,=1$$ gives each term to be $(1-1)=0$ implying the limit to be 0. Please explain to me where I went wrong.

Answer 1

You have a sum of the form $\sum_{k=1}^{n} a(k,n),$ where for each fixed $k,$ $\lim _{n\to \infty}a(k,n) = 0.$ Does that imply $\lim _{n\to \infty}\sum_{k=1}^{n} a(k,n) =0?$ Certainly not. Almost any Riemann sum situation is a counterexample. For example, $\sum_{k=1}^{n} k/n^2\to 1/2.$