Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges

I know that I must use the Limit Comparison test and my instinct tells me that this series will converge. I cannot, however, think of another series $b_k$, to compare my given series to in order to show that it converges.

Any hints?

Thank you!

  • $\begingroup$ Tip: Use \sum rather than \Sigma next time. $\endgroup$ – Ali Caglayan Oct 19 '14 at 15:13
  • $\begingroup$ @Alizter Thanks! Will do :) $\endgroup$ – DJS Oct 19 '14 at 15:15
  • $\begingroup$ It sums to $\zeta'(2)$, which apparently equals $-\frac{\pi^2}{6}(\gamma+12\zeta'(-1)+\ln2+\ln\pi-1)$ somehow. $\endgroup$ – Akiva Weinberger Oct 19 '14 at 15:38


$$\sum_{k=1}^{\infty} \frac{\text{ln}\ k}{k^2} < \int_{1}^{\infty}\frac{\text{ln}\ x}{x^2} dx = 1 $$

  • Try looking at the maximum value of the function $\displaystyle f(x) = \frac{\ln(x)}{\sqrt{x}}$ on $(1,\infty)$

  • $\displaystyle f'(x)= \frac{2-\ln(x)}{2x^{3/2}}$

  • $f'(x) <0$ for all $x>e^{2}$. Hence $f(x) \leq f(e^{2})$ for all $x\geq 1$. Hence $\displaystyle \frac{\ln(x)}{\sqrt{x}} \leq \frac{2}{e}$.

  • Now $\displaystyle \sum_{k=1}^{\infty} \frac{\ln{(k)}}{k^{2}} = \sum\limits_{k=1}^{\infty} \frac{\ln(k)}{\sqrt{k}} \times \frac{1}{k\sqrt{k}} \leq \frac{2}{e} \sum_{k=1}^{\infty} \frac{1}{k^{3/2}} <\infty$


Equals the constant in http://oeis.org/A073002 , which gives further links.


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