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

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

• The key-word here is "Riemann sum". What is wrong in your answer is that $$\left( \frac{1}{e^\frac{1}{n}} + \frac{1}{e^\frac{2}{n}} + \frac{1}{e^{\frac{3}{n}}} + \cdots + \frac{1}{e^\frac{n}{n}} \right)$$ is not equal to $$\frac{1}{e^\frac{1}{n} -1}$$ Jul 16, 2021 at 13:31
• @Gary this is from an exam intended at high schoolers. I am not familiar with this concept. Jul 16, 2021 at 13:35
• What if you write $$\sum_{k=1}^n \frac{e^{\frac{-k}{n}}}{n}=\frac 1 n \sum_{k=1}^n \Big[e^{-\frac 1n}\Big]^k$$ Jul 16, 2021 at 13:39
• Use $$\sum\limits_{k = 0}^n {ar^k } = a\frac{{r^{n + 1} - 1}}{{r - 1}}$$ for $r \neq 1$.
– Gary
Jul 16, 2021 at 13:58

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

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