# Bounding Coefficients of Dirichlet Series

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 $$O(n^{\frac{1}{\log \log n}})$$

• 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 – Gerry Myerson Aug 8 '19 at 0:14
• I gave a bound below for every $p\geq 2$, including non-integral $p$'s. – 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 $$1$$.