# Prove the asymptotic for the sum

I need to prove, that $$\sum_{k=1}^n{k^p\ln^qk} \sim \frac{n^{p+1}}{p+1}\ln^qn$$ where $p > -1$. I tried to use Stolz–Cesàro theorem to show that: $$\lim_{n\to \infty}\frac{n^p\ln^qn}{\frac{n^{p+1}}{p+1}\ln^qn - \frac{(n-1)^{p+1}}{p+1}\ln^q(n-1)}=1$$but, actually, I cannot understand how to show that. For me it is not obvious how to show the correctness of this limit. The $\frac{n^{p+1}}{p+1}$ term tells me that I need to use the L'Hôpital's rule, but in this case the denominator doubles the number of terms and everything gets worse.

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Do you mean $n \rightarrow \infty$? –  Pragabhava Oct 9 '12 at 16:30
There's a difference between writing 5ln x and writing 5\ln x in $\TeX$: $5ln x$ versus $5\ln x$. The latter form automatically has proper spacing before and after $\ln$ and does not italicize it. It is standard usage. (I changed it in the posting.) –  Michael Hardy Oct 9 '12 at 17:15
Pragabhava, yes, sure, $n$. –  Kos Oct 10 '12 at 6:36

$$\int_1^n k^p \ln^q kdk=\frac{n^{p+1}}{p+1}\ln^q n - \int_1^n \frac{q}{p+1}k^{p}\ln^{q-1}k dk$$
and you can show the integral on the r.h.s. is of lower order than the first term on the right, either by iterating $q-1$ times, or by bounding $\int_1^nk^p\ln^{q-1}k\leq n\cdot n^p\ln^{q-1}n$, which again is lower order.