# A derivation of a Riemann zeta function identity

Originally published at 狗和留美者不得入内. You can comment here or there.

Yesterday, I saw the following Riemann zeta function identity:

$\displaystyle\sum_{n=1}^{\infty} \frac{\sigma_a(n)\sigma_b(n)}{n^s} = \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}$.

I took some time to try to derive it myself and to my great pleasure, I succeeded.

Eventually, I realized that it suffices to show that

$\{(dd_a, dd_b, d^2 n) : d_a | n, d_b | n : d, d_a, d_b, n \in \mathbb{Z}\}$

and

$\{(dd_a, dd_b, n) : dd_a d_b | n : d, d_a, d_b, n \in \mathbb{Z}\}$

are equal as multisets. As sets, they are both representations of the set of $3$-tuples of positive integers such that the third is a multiple of the least common multiple of the first two. In the latter one, the frequency of $(a,b,c)$ is the number of $d$ that divides both $a$ and $b$ such that $ab | cd$. In the other one, if we write $(a,b,c)$ as $(d_1 d_2 a', d_1 d_2 b', c)$ where $\mathrm{gcd}(a', b') = 1$, the $ab | cd$ condition equates to $d_1^2 d_2 a'b' | c$, which corresponds to the number of $d_1$ dividing $a$ and $b$ and such that $d_1^2 | c$ and with that, $d_2a', d_2b'$ both dividing $d_1^2 / c$, which is the frequency of $(a,b,c)$ via the former representation.

The coefficients $\{a_n\}$ of the Dirichlet series of the LHS of that identity can be decomposed as follows:

$a_n = \displaystyle\sum_{d^2 | n, d_a | \frac{n}{d^2}, d_b | \frac{n}{d^2}} (dd_a)^a (dd_b)^b$.

The coefficients $\{b_n\}$ of the Dirichlet series of the RHS of that identity are

$b_n = \displaystyle\sum_{dd_a d_b | n} (dd_a)^a (dd_b)^b$.

Observe how both are equivalent in that via the multiset equivalence proved above, $n$ determines the same multiset of $(dd_a, dd_b)$ for both and across that, the values of the same function $(dd_a)^a (dd_b)^b$ are summed. Hence the two series are equal.

• #### 中国姓氏的南北分布

有意思的是分布是很不均匀的，也就是说好多姓是南方比例远高于北方比例或相反。从这一点，你如果知道一个人的父母的姓氏，你就可以对他是北方人还是南方人有个有点统计显著的估计。 读者可以在包含大姓分布图的…

• #### Why I think Stanford professor James Landay's piece on computing in China in 2011 exposed his idiocy

I dittoed his blog post on my LiveJournal too, see https://gmachine1729.livejournal.com/175023.html. For the original, see…

• #### China Will Overtake the US in Computing…Maybe, Someday… by Stanford professor James Landay in 2011

Original at https://dubfuture.blogspot.com/2011/12/china-will-overtake-us-in.html. [note: the following is a rough draft — I appreciate…

• Post a new comment

#### Error

Anonymous comments are disabled in this journal

default userpic