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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0Pragabhava, yes, sure, $n$. – 2012-10-10
1 Answers
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The terms in the summand are increasing monotonically. Why not try to bound above and below by integrals? The integrals, up to fiddling are
$\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.