This note is part of the set Exploring the Divisor Function.

In this set, we aim to get a general form for this sum:

\[\sum _{n=1}^{\infty}\frac{d\left(kn\right)}{n^2}\]

Where \(k\) is a positive integer.

So, instead of giving out everything on a note, why not split it up into several problems so that everybody can try it out by themselves?

I will give a clue here, and then you can go ahead to solve the first problem of this set, slowly progressing to the last problem, where you will finally be able to find a general form of the sum. You may skip steps, because your approach might be better than mine. If you do have a better approach, do post it!

Here's the first clue:

If \(f(n)\) is completely multiplicative, that is \(f(ab)=f(a)f(b)\), then

\[f*f(n)=d(n)f(n)\]

\[\left[\sum _{n=1}^{ \infty}\frac{f\left(n\right)}{n^s}\right]^2=\sum _{n=1}^{\infty }\frac{f\left(n\right)d\left(n\right)}{n^s}\]

Where \(*\) is the Dirichlet Convolution

and \(d(n)\) counts the number of divisors n.

I would post the solutions for the problems soon.

## Comments

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TopNewestI cannot work it out for 4 as \(k\) Can you help? – Joel Yip · 8 months, 3 weeks ago

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If you want to consider k=4, generalise it to \(k=p^a\), where p is prime. I'll post a solution to part 3 of the set by tomorrow, where you can use it for \(k=4\).

You can use part 1 of this set as a clue to part 3, but you'll have to be more creative. – Julian Poon · 8 months, 3 weeks ago

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– Joel Yip · 8 months, 3 weeks ago

thanks!Log in to reply

– Julian Poon · 8 months, 3 weeks ago

I have posted a solution to part 3.Log in to reply

my mind is blown – Joel Yip · 8 months, 3 weeks ago

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