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.
Prove the asymptotic for the sum
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limits
asymptotics
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0Do you mean $n \rightarrow \infty$? – 2012-10-09
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0There'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.) – 2012-10-09
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0Pragabhava, yes, sure, $n$. – 2012-10-10