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I am trying to calculate the euler product of $\frac{1}{\gamma}(n)$ where $\gamma(n)$ is the number of divisors of n. So I have that:

$\displaystyle D_{\frac{1}{\gamma}(n)}=\prod_p \left ( 1+ \frac{\frac{1}{\gamma}(p)}{p^s}+\frac{\frac{1}{\gamma}(p^2)}{p^{2s}}+\ldots \right )= \prod_p \left ( 1+ \frac{1}{\gamma(p)p^s}+\frac{1}{\gamma(p^2)p^{2s}}+\ldots \right )= \prod_p \left ( 1+ \frac{1}{2p^s}+\frac{1}{3p^{2s}}+\ldots \right ) =\prod_p \left ( 1+ \sum_1^\infty \frac{1}{(k+1)p^{ks}} \right ) $

However I am supposed to get this in the form $\prod -p^s\log(1-p^{-s})$ which seems a long way off, have I made a mistake or am I missing something?

Thanks for the help

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I'm sorry I don't understand what is $\frac{1}{\gamma}(n)$ is it $\frac{1}{\gamma(n)}$? –  Grobber Apr 21 '13 at 12:09
    
@Grobber erm yeah, that is what I was assuming, in the question it is just written like that with no more explanation anywhere sorry. –  hmmmm Apr 21 '13 at 12:29
3  
This is almost done. Use the identity $$\log \frac{1}{1-z} = \sum_{k\ge 1}\frac{z^k}{k}.$$ –  Marko Riedel Apr 21 '13 at 19:37
    
@MarkoRiedel Thanks very much that obviously does it cheers –  hmmmm Apr 21 '13 at 20:14
    
@MarkoRiedel Why not make your comment an answer? –  Zander May 19 '13 at 13:57
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