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I would like to ask help for the following question:

Simplify $$\sum_{x=0}^{+\infty}{e^{-x/\mu}}$$

Will integration work here? Meaning, can I do it this way?

$$\sum_{x=0}^{+\infty}{e^{-x/\mu}}=\lim_{a\rightarrow+\infty}\sum_{x=0}^{a}{e^{-x/\mu}}=\int_0^{+\infty}e^{-x/\mu}dx$$

Thanks for the help.

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It's a geometric series, with common ratio $r = e^{-1/\mu}$. It converges if and only if $|r|<1$ (this includes the case where $r$ is complex). If it converges, the easiest way to remember its value is

$$ \frac{\mbox{[First Term In Series]}}{1-\mbox{[Common Ratio]}}. $$

That way you don't have to fuss about whether the series starts at $x=0$ or $x=1$ and details like that.

What you are doing with the integral might tell you whether or not the series converges, but it definitely won't give you the value of the series if it converges. With a geometric series like this you don't need the Integral Test (that's what you are attempting to use). I would save the Integral Test for problems where you need it. And still, the Integral Test can tell you whether a series converges but to my knowledge it never gives you the value of a convergent series.

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For $\vert a \vert < 1$, we have $$\sum_{x=0}^{\infty} a^x = \dfrac1{1-a}$$ Now set $a=e^{-1/\mu}$.

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  • $\begingroup$ You should probably add that $|a|<1$. $\endgroup$ – Mercy King Dec 6 '13 at 2:10
  • $\begingroup$ @Mercy Added, Your Honour. $\endgroup$ – user17762 Dec 6 '13 at 2:17
  • $\begingroup$ You are welcome your Highness! $\endgroup$ – Mercy King Dec 6 '13 at 2:26

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