I'm trying to show:

$$\log \prod_p\frac{1}{1-p^{-s}}=\sum_p\sum_{n=1}^\infty \frac{1}{np^{ns}}$$

Using the properties of the logarithm I get:

$$\log \prod_p\frac{1}{1-p^{-s}}=\sum_p\log\frac{1}{1-p^{-s}}=\sum_p(\log 1-\log (1-p^{-s}))=$$ $$=-\sum_p \log(1-p^{-s})=-\sum_p \log\bigg(\frac{p^s-1}{p^s}\bigg)$$

The log can be expanded more, but it seems to get me farther from the result.

I realize that the sum over $n$ comes from a geometric sum, but I don't see where to get it from, or how to get rid of the logarithm.

Thanks in advance for any help or tips!

  • $\begingroup$ Hint: try the Taylor series for $\log(1-x)$ and let $x=p^{-s}$ $\endgroup$
    – rikhavshah
    Commented Aug 29, 2016 at 4:39

1 Answer 1


Progress you made: $$\log\prod_p \frac{1}{1-p^{-s}}=-\sum_p\log(1-p^{-s})$$

Taylor series: $$\log(1-x)=\sum_{n=1}^{\infty}\frac{-x^n}{n}$$

Plug and chug: $$\log(1-p^{-s})=\sum_{n=1}^{\infty}\frac{-p^{-sn}}{n}$$

$$\log\prod_p \frac{1}{1-p^{-s}}=-\sum_p\sum_{n=1}^{\infty}\frac{-p^{-sn}}{n}=\sum_p\sum_{n=1}^{\infty}\frac{1}{np^{sn}}$$


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