# The convergence of this series: $\sum\limits_{n=2}^\infty {1\over n^{\log n}}$

I came across a problem on convergence of series and I did not get into any idea about this -- any help or hints ? $$\sum_{n=2}^\infty {1\over n^{\log n}}$$

What about $n^{\log(\log n)}$ ?

• Making a guess as to what you mean, Baby Rudin 3.28-3.29 would seem helpful. Mar 7, 2016 at 5:59
• To show divergence if $a\le 1$, it is enough (why?) to show divergence when $n=1$. For this use the Integral Test. Mar 7, 2016 at 6:11
• Hello can you make it clear that my question is about n power logn Mar 7, 2016 at 9:20
• For some $N$, $n>N\implies\log(n)>1$. For some other $N$, $n>N\implies\log(\log(n))>1$.
– user65203
Mar 7, 2016 at 9:24
• Can you edit the problem with n power to the logn please i am not able to do so Mar 7, 2016 at 9:25

For which $a$ does $\sum_{n=2}^\infty {1\over n^a}$ converge? For which $a$ does it diverge?
If this converges, then $\sum_{n=2}^\infty {1\over n^a\log n}$ will certainly converge (do you see why?).
$$\frac{2^n}{(2^n)^{\log2^n}}=\frac{2^n}{2^{n^2\log2}}=\frac1{2^{n\left(n\log2-1\right)}}\le\frac1{2^n}$$