3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

Hint:

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.