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Does the sequence $\sum \limits_{n=1}^{\infty} \frac{\log n}{n^2}$ converge absolutely?

I know that $\sum \limits_{n=1}^{\infty} \frac{1}{n^s}$ for $s >1 $ converges absolute. So is it possible to show the absolute convergence with that knowledge?

fragant
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    This sequence has positive terms, so it converges absolutely if and only if it is convergent... – Crostul Jan 19 '16 at 15:16
  • A useful fact to prove is that for any $\epsilon > 0$ then for large enough $n$ (depending on $\epsilon$) we have $\log(n) < n^\epsilon$ (or even better $\lim_{n\to\infty} \frac{\log(n)}{n^\epsilon} = 0$). In normal words: the logarithm grows slower towards $\infty$ than any power. You can use this to show that $\sum \frac{\log(n)}{n^p}$ converges for all $p>1$ (comparison test with the series you quote). – Winther Jan 19 '16 at 16:20

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Yes. This sum converges. You might try to write $$\frac{\ln n}{n^2}=\frac{\ln n}{\sqrt n} \cdot \frac{1}{n^\frac{3}{2}}.$$ Apply the fact that $\lim\limits_{n\to\infty} \frac{\ln n}{\sqrt n}=0$ and compare your series by the series $\displaystyle{\sum_{n=1}^\infty \frac{1}{n^\frac{3}{2}}}.$

Marko
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Yes it does converge absolutely, we have by the integral test, $$0\leq\sum \limits_{n=1}^{\infty} \frac{\log n}{n^2}\leq\int_{1}^{\infty} \frac{\log x}{x^2}dx=1.$$ (the latter integral is obtained by integration by parts)

Euler's student
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You can use Cauchy Condensation test. The series is convergent iff the following series is convergent.

$$ {1\over\log_2e}\sum_{n=1}^\infty{2^n\log_2 2^n\over 2^{2n}}={1\over\log_2e}\sum_{n=1}^\infty{n\over 2^n} $$ The series above converges by ratio test and sums to ${2\over\log_2e}$.

Community
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Jack's wasted life
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