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How to evaluate$$\sum ^{\infty}_{n=1} {e}^{-n}$$

without using the easy-formula.

We easily notice a pattern.

$$\begin{align} S_1 &= e^{-1} \\ S_2 &= e^{-2} + e^{-1} = \frac{1 + e}{e^2} \\ S_3 &= e^{-3} + e^{-2} + e^{-1} = \frac{1 + e^2 + e}{e^3} \\ S_4 &= e^{-4} + e^{-3} + e^{-2} + e^{-1} = \frac{1 + e + e^2 + e^3}{e^4} \\ &\vdots \\ S_n &= \frac{1 + \sum^{n}_{k=1} e^k}{e^n}\end{align}$$

This won't get us anywhere.

So how can we evaluate this infinite sum using partial $nth$ sum?

Thanks!

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    $\begingroup$ Without using the 'easy formula' basically means you have to derive the 'easy formula'. Multiply series by $e^{-1}$ and compare with original series. $\endgroup$
    – Winther
    Oct 25 '14 at 13:47
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    $\begingroup$ No matter if you use the well-known formula for geometric sereis or not, you cannot arrive at any other result that $\frac{1}{e-1}$. So maybe consider the difference between $\frac1{e-1}$ and the partial sums and see if you can find a proof that it tends to $0$? $\endgroup$ Oct 25 '14 at 13:52
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Hint:

How about noticing this pattern? \begin{align} S &= \sum ^{\infty}_{n=1} {e}^{-n}=e^{-1}+e^{-2}+e^{-3}+\ldots \\ \frac{S}{e} &= e^{-2}+e^{-3}+e^{-4}+\ldots \\ S-\frac{S}{e} &= e^{-1} \\ S\left(1-\frac{1}{e}\right) &= \frac{1}{e} \\ S &= \frac{1}{e-1} \end{align}

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$$S_3=r+\color{green}{r^2+r^3}$$ $$rS_3=\color{green}{r^2+r^3}+r^4$$ $$S_3-rS_3=r-r^4=r(1-r^3)$$ $$S_3=r\frac{1-r^3}{1-r}.$$

You can replace $r$ by any real and $3$ by any integer, and in the limit, for $|r|<1$, $$S_\infty=\frac r{1-r}.$$

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Well, $$ (e^{-1}-1) \sum_{n=1}^N e^{-n} = \sum_{n=1}^N e^{-n-1} - \sum_{n=1}^N e^{-n} =e^{-N-1}-e^{-1} \to -e^{-1} \\\implies \sum_{n=1}^\infty e^{-n} = \frac {e^{-1} }{1-e^{-1}} $$

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