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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).

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  • $\begingroup$ 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
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For any $a > 0$, we have

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

Hence,

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

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

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