Partial sums of powers of the divisor function

Question

It is easy to establish that $$\sum_{n\le x}\tau(n) \sim n\log n$$ How would one find good bounds on $$\sum_{n\le x} \tau(n)^k $$ for some $k > 0$

Answer 1

We have $$\underset{n\leq x}{\sum}\left(\tau\left(n\right)\right)^{k}=O_{k}\left(x\left(\log\left(x\right)\right)^{2k-1}\right).$$You can find the proof here http://planetmath.org/displaystylesumnlextaunaoaxlogx2a1forage0 .