Consider the exponentiated Riemann-Zeta function $\zeta(s)^p$. If it is represented as

$$\zeta(s)^p = \sum_{n=1}^\infty\frac{a_n}{n^s}$$

Is there any upper bound we can put on $|a_n|$ in terms of $n$ and $p$.

For example, not that when $p = 2$, we get the divisor function which can be bounded above by $O(n^{\frac{1}{\log \log n}})$

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    $\begingroup$ Don't you get the number of ways of writing $n$ as a product of $p$ integers? And isn't that a much-studied function, with estimates widely available? E.g., oeis.org/A007425 $\endgroup$ – Gerry Myerson Aug 8 '19 at 0:14
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    $\begingroup$ I gave a bound below for every $p\geq 2$, including non-integral $p$'s. $\endgroup$ – GH from MO Aug 8 '19 at 2:20

Let me rename $p$ to $z$, because $p$ usually stands for prime numbers in the subject. I will assume that $z\geq 2$, but I will not assume that $z$ is an integer. The Dirichlet coefficients of $\zeta(s)^z$ form a generalized divisor function: $$\zeta(s)^\nu=\sum_{n=1}^\infty\frac{\tau_z(n)}{n^s},\qquad \Re(s)>1.$$ The generalized divisor function is multiplicative, and on prime powers it is given by $$\tau_z(p^\nu)=\binom{z+\nu-1}{\nu}.$$ Using this formula and its Taylor generating series $(1-x)^{-z}$, it is not hard to prove that $$\tau_z(p^\nu)\leq\min\bigl((\nu+1)^z,z^\nu\bigr).$$ Then we can proceed as in the proof of Theorem 2 in Section I.5.2 of Tenenbaum's book "Introduction to analytic and probabilistic number theory" to see that, for any $t>1$, $$\tau_z(n)\leq\exp\left(tz(2+\log\log n)+\frac{\log z\cdot\log n}{\log t}\right).$$ Choosing $t>1$ such that $$t(\log t)^2=\frac{\log z\cdot\log n}{z(2+\log\log n)},\tag{1}$$ the above bound becomes $$\tau_z(n)\leq\exp\left(\frac{\log z\cdot\log n}{(\log t)^2}+\frac{\log z\cdot\log n}{\log t}\right).$$ However, for the above choice of $t$, $$\log t=\log\log n+O\bigl(\log z+\log\log\log n\bigr),\tag{2}$$ as can be seen by checking the cases $\log n\leq z$ and $\log n>z$ separately. Let us now assume for simplicity that $\log n>z^2$. Then $\log n>t(\log t)^2\gg(\log n)^{1/2}$ by $(1)$, hence also $\log t\asymp\log\log n$. We conclude that $$\tau_z(n)\leq n^{\frac{\log z}{\log\log n}\left(1+O\left(\frac{\log z+\log\log\log n}{\log\log n}\right)\right)},\qquad n>\exp(z^2).$$ The implied constants here are absolute, i.e., they do not depend on $z$.

Remark. One can cover the range $1\leq z<2$ similarly, but with more care. In particular, everthing before $(2)$ is valid in this range, but $(2)$ fails e.g. when $n$ is fixed and $z$ approaches $1$.

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  • $\begingroup$ Would you know how to give a bound on the partial sums as well $\endgroup$ – Halbort Aug 8 '19 at 19:31
  • $\begingroup$ @Halbort: Partial sums of the Dirichlet series? Upper or lower bounds? Please ask this in a separate question. $\endgroup$ – GH from MO Aug 8 '19 at 21:07

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