Convergence test of $\sum_{n=2}^{\infty} \frac {1}{n} (\frac {1}{\ln n})^{\frac {3}{2}} $ I need to test for convergence  $\sum_{n=2}^{\infty} \frac {1}{n} (\frac {1}{\ln\ n})^{\frac {3}{2}} $. I figured that the comparison test will probably be the best option but I have no idea what to compare it with.
Thank you for your help.
 A: hint: Do an integral test $\displaystyle \int_{2}^\infty \dfrac{dx}{x{(\ln x)}^{3/2}}$ comparison. The series converges iff the above improper integral converges, and this is a nice problem so handle.
A: We can use the Cauchy condensation test
$$ 0 \ \leq\ \sum_{n=1}^{\infty} f(n)\ \leq\ \sum_{n=0}^{\infty} 2^{n}f(2^{n})\ \leq\ 2\sum_{n=1}^{\infty} f(n)$$
that is
$$\sum_{n=2}^{\infty} \frac {1}{n} \left(\frac {1}{\ln\ n}\right)^{\frac {3}{2}}\le \sum_{n=2}^{\infty} \frac {2^n}{2^n} \left(\frac {1}{\ln\ 2^n}\right)^{\frac {3}{2}}=\sum_{n=2}^{\infty}\frac{1}{n^\frac32{(\ln 2)}^\frac32}$$
A: 
I thought it might be instructive to present an approach that relies on neither the integral test nor Cauchy's condensation test.  Rather we use only straightforward arithmetic, recognition of a telescoping series, and the inequality $\log\left(1+\frac1n\right)\ge \frac1{n+1}$.

Making use of the inequality $\log\left(1+\frac1n\right)\ge\frac1{n+1}$, we see that
$$\begin{align}
\frac2{\sqrt{\log(n)}}-\frac2{\sqrt{\log(n+1)}}&=\frac{\log\left(1+\frac1n\right)}{\sqrt{\log(n)\log(n+1)}\left(\sqrt{\log(n)}+\sqrt{\log(n+1)}  \right)}\\\\
&\ge \frac1{(n+1)\log^{3/2}(n+1)}\tag1
\end{align}$$
Summimg both sides of $(1)$ from $n=2$ to $n=N$ and recognizing the left-hand side as a telescoping series reveals
$$0<\sum_{n=3}^{N+1}\frac1{n\log^{3/2}(n)}\le \frac2{\sqrt{\log(2)}}-\frac2{\sqrt{\log(N+1)}}$$
whence letting $N\to \infty$ shows that the series of interest converges.
And we are done.
A: HINT: Use the integral test and let $u=ln(x), du={\frac {1}{x}dx}$
...
A: Using Ermakoff's test for $\sum_{n=2}^{\infty} \frac {1}{n} (\frac {1}{\ln\ n})^{\frac {3}{2}}$:
$$\lim_\limits{x\to\infty}\frac{e^xf(e^x)}{f(x)}=\lim_\limits{x\to\infty}\frac{e^x\cdot \frac{1}{e^x}\cdot \frac{1}{x^{3/2}}}{\frac{1}{x}\cdot\frac{1}{(\ln x)^{3/2}}}=\lim_\limits{x\to\infty}\frac{(\ln x)^{3/2}}{x^{1/2}}=0<1.$$
Hence, the series converges.
