# How to compute this limit related to series?

How to compute $\lim_{N\rightarrow+\infty}\frac{\ln^2N}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^{2\ln N-2}$? thank you.

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You shouldn't erase the question. –  Vÿska Aug 27 '12 at 1:11
What have you tried? Did you try to apply the same technique as in this problem? –  sos440 Aug 27 '12 at 3:58

Because $f(x) =x^{2 \ln(n) - 2}$ is increasing function of $x$ for $n \geqslant 3$, we have $$\int_0^n \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} = \sum_{m=0}^{n-1} \int_{m}^{m+1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} \lt \frac{1}{n} \sum_{m=0}^{n-1} \left( \frac{m+1}{n} \right)^{2 \ln(n) - 2}$$ Thus: $$\frac{\ln^2(n)}{n} \sum_{k=1}^{n-1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \gt \ln^2(n) \left( \int_0^n \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} - 1\right) = \ln^2(n) \left( \frac{1}{2 \ln(n) - 1} -\frac{1}{n} \right)$$ Now, $$\lim_{n \to \infty } \frac{\ln^2(n)}{n} \sum_{k=1}^{n-1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} > \lim_{n \to \infty} \ln^2(n) \left( \frac{1}{2 \ln(n) - 1} -\frac{1}{n} \right) = \infty$$