# $\sum_{k=1}^\infty\frac{\log k}{k^p}$ and $\sum_{k=1}^\infty \frac{1}{k^{\log k}}$ both converge

Show that each of the following series converges

1) $\displaystyle \sum_{k=1}^\infty\dfrac{\log k}{k^p}$, $p > 1$.

2) $\displaystyle \sum_{k=1}^\infty \dfrac{1}{k^{\log k}}$.

Can anyone please help me start this problems. I think I can use the comparison theorem. Thank you.

• You can use the integral test – Samrat Mukhopadhyay Feb 11 '15 at 5:35
• For the second one, notice that $\log k>1$ for $k\geq 3$. – Eoin Feb 11 '15 at 5:35
• for both of them? – use0402 Feb 11 '15 at 5:36
• The first series is the derivative of the Riemann $\zeta$ function. – Lucian Feb 11 '15 at 11:52

Hint $$\int_{3}^\infty \frac{\log x}{x^p}dx=\int_{\log 3}^\infty e^{-(p-1)z}z dz<\frac{1}{(p-1)^2}\\ \int_{1}^{\infty}\frac{1}{x^{\log x}}dx=\int_{0}^{\infty}ze^{-z^2}dz=\frac{1}{2}$$
• For integral test to be valid the function has to be non negative and decreasing on some $[N,\infty)$. since $\log n\ge 0$ for $n\ge 3$, I used $3$. And you can always change from base $10$ to base $e$ by a constant multiplication right? – Samrat Mukhopadhyay Feb 11 '15 at 5:48
The Cauchy Condensation Test works well in these kinds of problems: $$\sum_{k=1}^\infty\frac{\log(k)}{k^p}\iff\sum_{k=1}^\infty\dfrac{k\log(2)2^k}{2^{kp}}$$ and $$\sum_{k=1}^\infty\frac1{k^{\log(k)}}\iff\sum_{k=1}^\infty\frac{2^k}{2^{k^2\log(2)}}$$