Help evaluating $\lim_{n\to \infty} \sum_{k=1}^n \frac{e^{\frac{-k}{n}}}{n}$

Question

$$ \lim_{n\to \infty} \sum_{k=1}^n \frac{e^{\frac{-k}{n}}}{n} $$ is what I am asked to evaluate. My working:

\begin{align} &= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^n e^{\frac{-k}{n}} \\ &= \lim_{n\to \infty} \frac{1}{n} \left( \frac{1}{e^\frac{1}{n}} + \frac{1}{e^\frac{2}{n}} + \frac{1}{e^{\frac{3}{n}}} + \cdots + \frac{1}{e^\frac{n}{n}} \right) \\ &= \lim_{n\to \infty} \frac{1}{n}\frac{1}{e^\frac{1}{n} -1} = \lim_{\frac{1}{n} \to 0} \frac{\frac{1}{n}}{e^\frac{1}{n} -1} \\ &= 1. \end{align}

Wrong. The answer is $1-\frac{1}{e}$. Where did I go wrong here?

Answer 1

The equality

$$\left( \frac{1}{e^\frac{1}{n}} + \frac{1}{e^\frac{2}{n}} + \frac{1}{e^{\frac{3}{n}}} + \cdots + \frac{1}{e^\frac{n}{n}} \right)=\frac{1}{e^\frac{1}{n} -1}$$

is false. I think what you want to see is a geometric sum

$$\sum_{k=1}^n e^{-k/n} = \sum_{k=1}^n \left(e^{-1/n}\right)^k =\frac{e^{-1/n-1}-1}{e^{-1/n}-1} $$

where the numerator takes the form

$$\lim_{n\to\infty}e^{-1/n-1}-1=\frac{1}{e}-1$$

and by L'Hopital's rule the denominator becomes

$$\lim_{n\to\infty} n \left (e^{-1/n}-1 \right ) = \lim_{n\to\infty}\frac{e^{-1/n}-1}{1/n}=-1$$