This is Exercise 4.3.4(a) in Montgomery and Vaughn “Multiplicative Number Theory…”

Let $ f_1$ , $ f_2$ be totally multiplicative functions with $ |f_i(n)| \leq 1$ . Show that for Re$ (s)> 1$ , $ $ \left(\sum_{n \geq 1} n^{-s} \left(\sum_{d \mid n} f_1(d) \right)\left(\sum_{d \mid n} f_2(d) \right) \right) \left(\sum_{n \geq 1} n^{-2s} f_1(n)f_2(n) \right)$ $ $ $ = \zeta(s)\left(\sum_{n \geq 1} n^{-s} f_1(n) \right)\left(\sum_{n \geq 1} n^{-s} f_2(n) \right)\left(\sum_{n \geq 1} n^{-s} f_1(n)f_2(n) \right).$ $

This contains Ramanujan’s $ \sum n^{-s} \sigma_a(n)\sigma_b(n) = \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}$ as a special case.

I would like to know if there’s a simpler proof of the above general result other than what I did.

After some manipulation, the LHS is $ $ \sum_{n \geq 1} n^{-s} \sum_{d^2 \mid n} \left(\sum_{e \ d \mid e \mid \frac{n}{d}} f_1(e)\right)\left(\sum_{e \ d \mid e \mid \frac{n}{d}} f_2(e)\right).$ $

After some manipulation, the RHS is $ $ \sum_{n \geq 1} n^{-s} \sum_{d \mid n} f_2\left(\frac{n}{d}\right)\left(\sum_{e \mid d} f_1(e) \right)\left(\sum_{e \mid \frac{n}{d}} f_1(e) \right).$ $

I was unable to show directly that these coefficients matched up, so using the fact that both sides are multiplicative functions of $ n$ (the convolution of (totally) multiplicative functions is multiplicative), I checked that both sides matched on prime powers. This was straightforward, but tedious.

Is there a simpler way?

I was also unable to get anywhere by looking at the Euler products.

Written with convolution notation, $ f*g(n) := \sum_{d \mid n} f(d)g \left(\frac{n}{d} \right)$ , the identity is the following complicated thing.

$ $ \left((f_1 * 1)(f_2 * 1) \right) * (f_1 \circ \sqrt{.})(f_2 \circ \sqrt{.})1_{n \ \text{-square}} = 1 * f_1 * f_2 * f_1f_2.$ $

Perhaps $ f_i^{-1}= \mu f_i$ might be helpful here.

At the very least, it would be nice if someone could point me in the direction of simple proofs of Ramanujan’s identity above. (I verified that also by looking at prime powers, and it wasn’t very insightful.)