Integral approach for infinite sum of $e^{-n}$

Question

A while ago a posted this same problem, I have a different approach, just need a little help...

$$\displaystyle e^{-n} = \int_{0}^{e} -nx^{-(n+1)} \,dx$$

Originally, we had, $\displaystyle \sum_{n=1}^{\infty} e^{-n}$

This is:

$$\displaystyle \int_{0}^{e} \sum_{n=1}^{\infty} -nx^{-(n+1)} \,dx$$

\begin{align} S_1 &= -x^{-2} \\S_2 &= -x^{-2} - 2x^{-3} \\S_3 &= -x^{-2} - 2x^{-3} - 3x^{-4} \\S_4 &= -x^{-2} - 2x^{-3} - 3x^{-4} - 4x^{-5} \end{align} If someone can help me find the sum, I can finish this problem.

Answer 1

The simplest solution is probably to compute directly the sum: $$ \sum_{n=1}^N e^{-n} = e^{-1}\frac {1-e^{-N}}{1-e^{-1}} $$

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NB: This comes from \begin{align}(1-e^{-1}) \sum_{n=1}^N e^{-n} &= \sum_{n=1}^N e^{-n}e^{-1} - \sum_{n=1}^N e^{-n} \\&= \sum_{n=2}^{N+1} e^{-n} - \sum_{n=1}^N e^{-n} \\&= e^{-1} - e^{-(N+1)} \end{align}