Upbounds for Ramanujan $\tau(n)$ function

Question

For prime $p$, $|\tau(p)|\le 2p^{11/2}$.

I am looking for the upbound for $|\tau(n)|$,$n\in \mathbb{N}$.

Thanks- mike

Answer 1

Since $\tau(n)$ is a multiplicative function, i.e., satisfies $\tau(nm)=\tau(n)\tau(m)$ for all $m,n$ with $gcd(m,n)=1$, we obtain an upper bound simply by multiplying the bound for primes. If $n=p_1^{e_1}\cdots p_r^{e_r}$. Then $|\tau(n)|=|\tau(p_1)|^{e_1}\cdots |\tau(p_r)|^{e_r}\le 2^r(p_1^{\frac{11}{2}e_1}\cdots p_r^{\frac{11}{2}e_r})$.

Furthermore one can show that $$ |\tau(n)|≤n^{\frac{11}{2}}(\log (n))^{−1/2+o(1)} $$ for a set of natural numbers $n$ of asymptotic density $1$. For details see here. This is better than the usual estimate $$ |\tau(n)|\le d(n)n^{\frac{11}{2}}\le n^{\frac{11}{2}}(\log (n))^{\log(2)+o(1)} $$ which holds for almost all $n$. Here $d(n)$ denotes the number of divisors of $n$.