Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 11H / d / Solution

If $n={\prod }_{p^{\alpha }\parallel n}{p}^{\alpha }$, multiplicativity and the prime-power formula for the divisor function giveChoose $P_0=2^{1/ϵ}$. For $p\ge P_0$ and $\alpha \ge 1$, the elementary bound $\alpha +1\le 2^{\alpha }$ givesFor each of the finitely many primes $p<P_0$, the sequence $(\alpha +1)p^{-\alpha ϵ}$ tends to zero and therefore has a finite maximum $M_p$ over $\alpha \ge 0$. Consequentlyuniformly in $n$, which proves $\tau (n)\le C(ϵ)n^{ϵ}$.