Does $\sum_{n=2}^ \infty \frac 1 {n \sqrt {\ln n}}$ converge? I want to figure out if this sum converges or diverges: $$\sum_{n=2}^ \infty \frac 1 {n \sqrt {\ln n}}$$
I tried comparing it to the harmonic series, but this is less than that so it was no use. The limit comparison test with the harmonic series doesn't seem to work either, as it gives $\infty$ or $0$. I thought of using the Integral Test, but this doesn't seem to have an obvious integral as far as I can tell. 
How should this be done?
 A: By the Cauchy condensation test, your series is convergent iff
$$ \sum_{n\geq 2}\frac{1}{\sqrt{n}} $$
is convergent, but obviously that is not the case.
A: With $u = \ln(t)$:
$$\int_{2}^{\infty}\frac{1}{t\sqrt{\ln(t)}}dt = \int_{\ln(2)}^{\infty}\frac{du}{\sqrt{u}} = \int_{\ln(2)}^{\infty}\frac{du}{u^{\frac12}}$$
This diverges because the power is $\frac12 \le 1$.
By integral test, the series diverges.
A: Observe that $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n \ln n}= \infty$, since $\displaystyle \frac{d}{dx}[\ln(\ln(x))]=\frac{1}{x \ln x}$. 
Notice that $\displaystyle \frac{1}{n\sqrt{\ln n}} > \frac{1}{n \ln n} $, hence $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\sqrt{\ln n}}= \infty $.
A: I thought that it might be of interest to present an approach that relies on pre-calculus tools only.  To that end we proceed.
Note that we have
$$\begin{align}
\sqrt{\log(N+1)}&=\sum_{n=1}^N \left(\sqrt{\log(n+1)}-\sqrt{\log(n)}\right)\\\\
&=\sum_{n=1}^N \frac{\log\left(1+\frac1n\right)}{\sqrt{\log(n+1)}+\sqrt{\log(n)}}\\\\
&\le \sum_{n=1}^N \frac1{2n\sqrt{\log(n)}}
\end{align}$$
Inasmuch as $\lim_{N\to \infty}\log(N+1)= \infty$, the series of interest diverges.
