For the first series, it is obvious that it is an alternating series in the form $\sum (-1)^{k-1} u_k$, where $u_k$ tends to zero monotonically, so it does converge, but I'm having trouble proving that $\frac{\log k}{\sqrt{k}}$ tends to zero monotonically.

Intuitively, I know that $\sqrt{k}$ grows much faster than $\log k$, so it makes sense, but I'm at a loss for how to prove it rigorously.

The second series is:

$$\sum a_k \;\;\mbox{where} \;\;a_k = \begin{cases} \frac{1}{k^2} & \text{if} \;k\; \text{is odd} \\ -\frac{\log k}{k^2} & \text{if} \;k\; \text{is even} \end{cases} $$

My line of thought is that $\sum \frac{1}{k^2}$ converges (well known fact), and $\sum \frac{\log k}{k^2}$ converges, but I'm not sure how to prove it. If I could prove that $\sum \frac{\log k}{k^2}$ converges, then the whole series obviously does converge then.

Any help would be appreciated (or easier methods).

  • $\begingroup$ Why not use de l'Hospital theorem to prove $\frac{\log k}{\sqrt{k}}$ tends to zero as k tends to infinity? $\endgroup$ – john melon Jan 5 '16 at 18:17
  • $\begingroup$ For $ k > 1, \log k > 0,$ and $$k = \exp \log k = 1 + \frac{\log k}{1} + \frac{(\log k)^2}{2!} + \frac{ (\log k)^3}{3!} + \dots +$$, so, $(\log k)^3/6 \leq k$ and $ \log k \leq \sqrt[3]{6k}.$ $\endgroup$ – Arin Chaudhuri Jan 26 '16 at 18:39

For any $a > 0$, we have

$$ \log k = \frac{\log k^a}{a} \leqslant \frac{k^a}{a}.$$


$$\frac{\log k}{k^2} < \frac{2k^{1/2}}{k^2}= \frac{2}{k^{3/2}}.$$

By the comparison test, $\displaystyle \sum \frac{\log k}{k^2}$ converges.


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.