# My book states that $\sum_{n=1}^{\infty}r^{-n} = \frac{1}{r-1}$ for $r > 1$

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 $0$. – Brian M. Scott Jul 19 '12 at 0:33

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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Ohhh gosh ok that clears things up, thanks. – Thoth Jul 19 '12 at 0:39

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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thankssssssssss – Thoth Jul 19 '12 at 0:40

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.

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