Let $\tau\_k(n)$ count the number of ways of representing $n$ as the product of $k$ natural numbers. It is known that:


$$D\_k(x) = \sum\_{n \leq x} \tau\_k(n) = xP\_k(\log x) + O(x ^{1 - \frac{1}{k-1}}(\log x)^{k-2}), \; \forall k \geq 2$$


Where $P\_k$ is a polynomial of degree $k-1$ with leading coefficient $\frac{1}{(k-1)!}$


I am asked to prove this. We may assume the base case as it is a well known result. 


The assuming the result for all $l \leq k$:


$$D\_k(x) = \sum\_{mn \leq x} \tau\_{k-1}(n) = \sum\_{n\leq x}\lfloor{\frac{x}{n}}\rfloor\tau\_{k-1}(n)$$


$$= \sum\_{n \leq x}(\frac{x}{n} + O(1))\tau\_{k-1}(n) = x\sum\_{n \leq x} \frac{\tau\_{k-1}(n)}{n} + O(\sum\_{n\leq x}\tau\_{k-1}(n))$$


Using Abel's Summation formula, we may see that:


$$\sum\_{n \leq x} \frac{\tau\_{k-1}(n)}{n} = P\_k(\log x) + O(x^{\frac{-1}{k-1}}(\log x)^{k-3})$$


So:


$$D\_k(x) = xP\_k(\log x) + O(x^{1 - \frac{1}{k-1}}(\log x)^{k-3}) + O(\sum\_{n \leq x}\tau\_{k-1}(n))$$


Using the induction hypothesis:


$$O(\sum\_{n \leq x}\tau\_{k-1}(n)) = O(xP\_{k-1}(\log x) + O(x^{1 - \frac{1}{k-1}}(\log x)^{k-3}))$$


$$= O(x(\log x)^{k-2}) + O(x^{1 - \frac{1}{k-1}}(\log x)^{k-3})) = O(x(\log x)^{k-2})$$


Thus we get:


$$D\_k(x) = xP\_k(\log x) + O(x^{1- \frac{1}{k-1}}(\log x)^{k-3}) + O(x(\log x)^{k-2})$$


$$= xP\_k(\log x)+O(x(\log x)^{k-2})$$


However, this error term is too large. How can I go about reducing it?



Piltz Divisor Problem
You've discovered some of the primary motivation for the invention of the [Dirichlet hyperbola method](http://www.maths.manchester.ac.uk/~mdc/MATH31022/2011-12/notes/Notes5%202013%20Sums%20of%20Arithmetic%20functions%20level%204.pdf). Since $\tau\_k = \tau\_{k-1}\*1$ (which you are already using), you can take advantage of the extra parameter ($U$ and $V$ in the linked document, instead of just $x$) to get a better error term—just as in the proof of the base case $k=2$.


