1
$\begingroup$

I read the note, where the lecturer prove $$ \log\zeta(s)=\sum_{n=2}^{\infty}\frac{\Lambda(n)}{\log n}\frac{1}{n^s} $$ for all $\sigma >1$. The only step that I don't understand is the following equation $$ \sum_{p}\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{p^{ks}}=\sum_{n=2}^{\infty}\frac{\Lambda (n)}{\log n}\frac{1}{n^s}. $$ I asked the lecturer, how he deduced to rewrite from the left side to the right side. His answer was that he used Fundamental theorem of arithmetic. How did he do that, if there are two sum signs, to transform them into one sum sign? Note that by definition of Von Mangoldt function, $$ \frac{\Lambda (n)}{\log n}=\begin{cases} \frac{1}{k} & \text{ if } n=p^k \text{ for some $k\geq 1$ and prime $p$ } \\ 0 & \text{ otherwise.} \end{cases} $$

$\endgroup$
  • 2
    $\begingroup$ There is nothing to prove. $\sum_p$ means summing over the prime numbers. $\sum_p \sum_{k \ge 1} \frac{(p^k)^{-s}}{k} = \sum_{n=1}^\infty n^{-s} a(n)$ is a Dirichlet series. What is the value of those coefficients $a(n)$ ? $\endgroup$ – reuns Dec 13 '16 at 18:30
1
$\begingroup$

Note that $\Lambda(n) = 0$ unless $n$ is equal to a prime power, so that $n = p^k$ for some prime $p$ and some integer $k \geq 1$. This means that we can delete all of the terms in the sum for which $n$ is not of this form (because these terms are $0$), so that \[\sum_{n = 2}^{\infty} \frac{\Lambda(n)}{n^s \log n} = \sum_{\substack{n = 2 \\ n = p^k \text{ for some prime $p$ and $k \geq 1$}}}^{\infty} \frac{\Lambda(n)}{n^s \log n}.\] If $n = p^k$, then $\Lambda(n) = \log p$, so that this is \[\sum_{\substack{n = 2 \\ n = p^k \text{ for some prime $p$ and $k \geq 1$}}}^{\infty} \frac{\log p}{(p^k)^s \log p^k} = \sum_{\substack{n = 2 \\ n = p^k \text{ for some prime $p$ and $k \geq 1$}}}^{\infty} \frac{1}{k p^{ks}}\] since $(p^k)^s = p^{ks}$ and $\log p^k = k \log p$.

Finally, the sum over all integers $n \geq 2$ of the form $p^k$ for some prime $p$ and $k \geq 1$ is the same as the sum over all primes $p$ and all integers $k \geq 1$ (that is, there is a bijection between these two sets), so this is equal to \[\sum_p \sum_{k = 1}^{\infty} \frac{1}{k p^{ks}}.\]

$\endgroup$
  • $\begingroup$ Many thanks for your answer. Now I also read it. $\endgroup$ – user243301 Dec 14 '16 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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