6
$\begingroup$

I would appreciate a hint concerning how to surpass the roadblock I've encountered in my attempt at a proof below. A nicer proof than mine would also help (Edit: The latter part is now done by Gerry Myserson, the prior remains).

Attempt at a proof (below):

As $\sigma_a(x)$ is completely multiplicative, we can take the infinite product of prime series: $$\sum_{n=1}^\infty \frac{\sigma_a(n)}{n^s}= \prod_{\text{p prime}}\sum_{k=0}^\infty \frac{\sigma_a(p^k)}{p^{ks}}$$ $$=\prod_{\text{p prime}}\sum_{k=0}^\infty \frac{\frac{p^{(k+1)a}-1}{p^a-1}}{p^{ks}}$$ $$=\prod_{\text{p prime}}\sum_{k=0}^\infty \frac{1}{p^a-1}\left[\frac{p^{(k+1)a}}{p^{ks}}-\frac{1}{p^{ks}}\right]$$ $$=\prod_{\text{p prime}} \frac{1}{p^a-1}\left[p^a\zeta(s-a)-\zeta(s)\right]$$ $$=\zeta(a)\prod_{\text{p prime}} \left[\zeta(s-a)-p^{-a}\zeta(s)\right]$$ I cannot see how to extract $\zeta(s)\zeta(s-a)$ from this.

$\endgroup$
2
$\begingroup$

Since $\sigma_a=f\ast g$ is the Dirichlet product with $f(n)=1$ and $g(n)=n^a$, and since multiplication of Dirichlet series is given with this product, we obtain $$ \sum_{n=1}^{\infty}\sigma_a(n)n^{-s}=\sum_{n=1}^{\infty}n^{-s}\sum_{n=1}^{\infty}n^an^{-s}=\zeta(s)\zeta(s-a). $$

$\endgroup$
1
$\begingroup$

$$\zeta(s)\zeta(s-a)=\sum_j(1/j^s)\sum_k(k^a/k^s)=\sum_n c(n)/n^s$$ where we have to prove $c(n)=\sigma_a(n)$. But every factorization $n=jk$ contributes $k^a$ to the coefficient of $n^{-s}$.

$\endgroup$

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.