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

$$D_k(x) = sum_n leq x tau_k(n) = xP_k(log x) + O(x ^1 - frac1k-1(log x)^k-2), ; forall k geq 2$$

Where $P_k$ is a polynomial of degree $k-1$ with leading coefficient $frac1(k-1)!$

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 $l leq k$:

$$D_k(x) = sum_mn leq x tau_k-1(n) = sum_nleq xlfloorfracxnrfloortau_k-1(n)$$

$$= sum_n leq x(fracxn + O(1))tau_k-1(n) = xsum_n leq x fractau_k-1(n)n + O(sum_nleq xtau_k-1(n))$$

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

$$sum_n leq x fractau_k-1(n)n = P_k(log x) + O(x^frac-1k-1(log x)^k-3)$$

So:

$$D_k(x) = xP_k(log x) + O(x^1 - frac1k-1(log x)^k-3) + O(sum_n leq xtau_k-1(n))$$

Using the induction hypothesis:

$$O(sum_n leq xtau_k-1(n)) = O(xP_k-1(log x) + O(x^1 - frac1k-1(log x)^k-3))$$

$$= O(x(log x)^k-2) + O(x^1 - frac1k-1(log x)^k-3)) = O(x(log x)^k-2)$$

Thus we get:

$$D_k(x) = xP_k(log x) + O(x^1- frac1k-1(log x)^k-3) + O(x(log x)^k-2)$$

$$= xP_k(log x)+O(x(log x)^k-2)$$

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

You've discovered some of the primary motivation for the invention of the Dirichlet hyperbola method. 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 $k=2$.