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We have $\frac{\zeta(s-1)^2}{\zeta(s)} = \sum\limits_{n\ge 1} \frac{a_n}{n^s}$, where $a_n = \sum\limits_{d|n} \mu(d) \sigma_0(\frac{n}{d}) \frac{n}{d} = \sum\limits_{d|n} \phi(d) \frac{n}{d}$. Here $\zeta$ is the Riemann zeta, $\mu$ is the Mobius function, $\sigma_0$ is the number of divisors, and $\phi$ is the Euler totient. What is the best bound for the coefficients $a_n$? I have obtained a very rough bound $a_n \le n^3$.

This was originally related to a problem in field theory, but I realized I can trivially get a much better bound in that situation.

Vik78
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    You can do better than $n^3$ even trivially: At worst $\phi(d) = d - 1$ (for primes) and extra doubly worst take $\sum\limits_{d \mid n} \phi(d) n/d \leq \sum\limits_{d \leq n} n \frac{d - 1}{d} = n^2 - n \sum\limits_{d \leq n} \frac{1}{d}$. – Jakob Streipel Mar 03 '23 at 16:03
  • Thanks— yes I realized soon after this it was easy to improve on somewhat – Vik78 Mar 03 '23 at 16:20
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    Very slightly less trivially, but still nothing fancy: $n \sum\limits_{d \mid n} \frac{d - 1}{d} \leq n \sigma_0(n)$ since $\frac{d - 1}{d} \leq 1$. But $\sigma_0(n) \leq 2 \sqrt{n}$, so we get a bound of size $n^{3/2}$. – Jakob Streipel Mar 03 '23 at 16:29
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    Note also that the sum is $\sum\limits_{a = 1}^n \gcd(a, n)$, which people have asked about e.g. here. – Jakob Streipel Mar 03 '23 at 16:33

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Summarising the ideas in my comments.

We can get some trivial upper bounds by bounding $\phi(d) \leq d - 1$ trivially (as attained by primes): $$ \sum_{d \mid n} \phi(d) \frac{n}{d} \leq n \sum_{d \mid n} \frac{d - 1}{d}. $$ At this point we can even more trivially bound the sum over $d \mid n$ by the sum $d = 1, 2, \dots, n$ and get $$ \sum_{d \mid n} \phi(d) \frac{n}{d} \leq n^2 - n \sum_{d = 1}^n \frac{1}{d} $$ so we get an upper bound of $a_n \leq n^2$ essentially for free. (The bound $\frac{d - 1}{d} \leq 1$ gives this $n^2$ without a secondary term too, of course.)

We can do slightly better by using $\frac{d - 1}{d} \leq 1$ and instead counting divisors of $n$, meaning that any bound on $\sigma_0(n) = \sum\limits_{d \mid n} 1$ gives us a bound on our sum for free. E.g., the simplest version $\sigma_0(n) \leq 2 \sqrt{n}$ (since divisors come in pairs) gives $a_n \leq 2 n^{3/2}$.

Any bound on $\sigma_0(n)$ of course works, so using any bound like e.g. this one will give a substantial saving.

Therefore, to have any hope of doing better than something of the quality $n^{1 + \varepsilon}$ (or $n$ times some power of $\log$) we'd need to further analyse $\sum\limits_{d \mid n} \phi(d) / d$. This won't save us very much though, since (as analysed in this answer) for $n = \prod\limits_i p_i^{k_i}$ we have $$ \sum_{d \mid n} \frac{\phi(d)}{d} = \prod_i \Bigl( 1 + k_i \frac{p_i - 1}{p_i} \Bigr). $$ To get a sense of how little this could possibly saves us, recall how $$ \sigma_0(n) = \sum_{d \mid n} 1 = \prod_i (1 + k_i). $$ Hence using the bound $\phi(d) / d \leq \frac{d - 1}{d} \leq 1$ is not all that bad, which heuristically is borne out by the sense in which $\phi(n)$ is "always 'nearly $n$'."

Jakob Streipel
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