Problem with tau function

Question

Show that $\tau(n)^a$ is $o(n)$ (small $o$) for all $a$ real, with $\tau(n)$ the tau function (number of divisors of n).

I have not completed the demonstration of this.

Answer 1

Not widely known, it is a result of Nicolas and Robin that $$ \tau(n) \leq n^{\left( \frac{1.066018678...}{\log \log n} \right)} $$ with the constant $1.066...$ chosen for equality at $$ n= 6983776800, \; \; \tau(n) = 2304 $$ and nowhere else.

Canadian Mathematical Bulletin, volume 26, number 4, December 1983, pages 485-92: Majorations explicites pour le nombre de diviseurs de $N.$

There are also three slightly stronger results in Robin's 1983 dissertation at Limoge. With equality at a number $n$ near $6.929 \cdot 10^{40},$ $$ \tau(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.934850967971...}{\log \log n} \right)} $$ Note that this gives a qualitative version of Wigert's theorem, Theorem 317 in Hardy and Wright, page 262 in the fifth edition. But Wigert did not just give an upper bound, he also said that $\limsup$ of the appropriate expression is $\log 2.$

well, see https://mathoverflow.net/questions/43103/what-is-the-lower-bound-for-highly-composite-numbers/43105#43105