Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
9  
The summation is from $1$, not from $0$. –  Brian M. Scott Jul 19 '12 at 0:33

3 Answers 3

up vote 5 down vote accepted

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 $$\sum_{n} g_n = {\hbox{first term}\over 1 - \hbox{common ratio}}$$

share|improve this answer
    
Ohhh gosh ok that clears things up, thanks. –  cactuar 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 $\frac{1}{r-1}$.

share|improve this answer
    
thankssssssssss –  cactuar 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.