# Convergence of series of positive terms: $\sum_{n=2}^{\infty}\frac{1}{{(\log n)}^{p}}$

I have applied Cauchy condensation test for to test the convergence the series $\sum_{n=2}^{\infty}\frac{1}{{(\log n)}^{p}}$, where p is constant, I got $\frac{1}{{\log 2}^{p}}\sum_{k=1}^{\infty}\frac{2^{k}}{k^{p}}$ . I do not understand for which value of p such that the original series is convergent. Also have used Cauchy integral test but did not solve the improper integral $\int_{2}^{\infty} {\frac{1}{(\log{x})^{p}}}dx$. I do not understand the convergent or not, if it convergent what is the value of p will be. Please some one help me. Thanks

Hint. Using the Cauchy condensation test, you are led to consider convergence/divergence of the series $$\sum_{k=1}^{\infty}\frac{2^{k}}{k^{p}}.$$ But, for any fixed $p$, we have $$\lim_{k\to\infty}\frac{2^{k}}{k^{p}} \neq 0$$ thus your inital series is always divergent.
The integral test should work too. Setting $logx=t$ gives $dx=e^tdt$ gives a new integral of the form $\frac{e^t}{t^n}$ integrating to infinity. I see an e-power in the numerator against a polynomial term in the denominator so I am getting "wet feet" here. Can you finish it?