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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.

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  • $\begingroup$ You can use the integral test $\endgroup$ – Samrat Mukhopadhyay Feb 11 '15 at 5:35
  • $\begingroup$ For the second one, notice that $\log k>1$ for $k\geq 3$. $\endgroup$ – Eoin Feb 11 '15 at 5:35
  • $\begingroup$ for both of them? $\endgroup$ – use0402 Feb 11 '15 at 5:36
  • $\begingroup$ The first series is the derivative of the Riemann $\zeta$ function. $\endgroup$ – Lucian Feb 11 '15 at 11:52
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Hint $$\int_{3}^\infty \frac{\log x}{x^p}dx=\int_{\log 3}^\infty e^{-(p-1)z}z dz<\frac{1}{(p-1)^2}\\ \int_{1}^{\infty}\frac{1}{x^{\log x}}dx=\int_{0}^{\infty}ze^{-z^2}dz=\frac{1}{2}$$

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  • $\begingroup$ 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? $\endgroup$ – Samrat Mukhopadhyay Feb 11 '15 at 5:48
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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 $$ \sum_{k=1}^\infty\frac1{k^{\log(k)}}\iff\sum_{k=1}^\infty\frac{2^k}{2^{k^2\log(2)}} $$

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