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On page 9 of Edwards' Riemann's Zeta Function, he uses the equality $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$ to prove an identity connecting the gamma function and the Riemann zeta function. But how can this equality be right? Taking the inverse of $r$ we obtain:

$\sum_{n=1}^{\infty}(\frac{1}{r})^n = \frac{1}{1-\frac{1}{r}} = \frac{r}{r-1} \neq \frac{1}{r-1}$

However, I double checked the identity (third down) on wikipedia and in the book it's correct as stated, thus from my, apparently erroneous, perspective Edwards derives the correct formula from the incorrect application of the geometric series summation, thus I must be missing something, where have I gone wrong?

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    The summation is from $1$, not from $0$.2012-07-19

3 Answers 3

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If $r > 1$, you have $\sum_{n=1}^\infty r^{-n} = {1/r\over 1 - 1/r} = {1\over r - 1}.$ However, $\sum_{n=0}^\infty r^{-n} = {1\over 1 - 1/r} = {r\over r - 1}.$ Remember for a geometric series $\sum_n g_n$ that decays to zero, you have $\sum_{n} g_n = {\hbox{first term}\over 1 - \hbox{common ratio}}$

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The summing is from $1$ to infinity. The first term of your geometric series is $\frac{1}{r}$. So your sum should be $\frac{1}{r}\frac{1}{1-\frac{1}{r}}$. This is indeed $\frac{1}{r-1}$.

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You have $\sum_{n=1}^{\infty}r^{-n}=S$, then $rS=\sum^{\infty}_{n=0}r^{-n}$. The later sum can be easily evaluated to be $\frac{1}{1-\frac{1}{r}}$ by considering geometric series. And you get $\frac{1}{r-1}$ as claimed.