Examine the convergence of $\sum_{n=1}^\infty \frac1{n(\log_2n)}$ 
Examine the convergence of $\displaystyle \sum_{n=1}^\infty \frac1{n(\log_2n)}$

My Work:
Make $x_n=\frac1{n(\log_2n)}$, then $\lim_{n\to\infty} x_n$ must be equal to zero if this series converges. It is equal to zero as the denominator will approach infinity as n approaches infinity.
Neither the root test, nor the ratio test appears to be useful in this scenario. The series is also not geometric. I am currently trying to figure out a comparison test that I can use but nothing is coming to mind. Any help on this problem would be greatly appreciated.
Just some background information, this is one of my practice problems for an upcoming exam in my analysis course.
 A: Key Point : Use Cauchy's condensation test.
Let , $\displaystyle f(n)=\frac{1}{n\log_2 n}=\frac{\log_e 2}{n\log_e n}$. Let $a(>1)$ be a real number. Then , $$a^nf(a^n)=\frac{a^n \log 2}{a^n n\log a}=\frac{\log 2}{\log a}.\frac{1}{n}.$$As $\displaystyle \sum \frac{1}{n}$ is divergent so $\displaystyle \sum a^nf(a^n)$ is divergent and hence series $\displaystyle \sum f(n)$ is divergent.
A: HINT:
First, write $\log_2n=\frac{\log n}{\log 2}$.  Then, note that since $\frac{\log 2}{n\log n}$ is strictly decreasing for $n\ge 2$, we have for all $N\ge 2$ the inequality
$$\sum_{n=2}^{N}\frac{1}{n\log_2 n}=\sum_{n=2}^{N}\frac{\log 2}{n\log n}\ge\int_2^{N+1}\frac{\log 2}{x\log x}\,dx$$
A: Divergence
Using the inequality
$$
\log(1+x)\le x\tag{1}
$$
twice, we get
$$
\begin{align}
\log(\log(n+1))
&=\log\left(\log(n)+\log\left(1+\frac1n\right)\right)\\
&=\log(\log(n))+\log\left(1+\frac{\log\left(1+\frac1n\right)}{\log(n)}\right)\\
&\le\log(\log(n))+\frac{\log\left(1+\frac1n\right)}{\log(n)}\\
&\le\log(\log(n))+\frac1{n\log(n)}\tag{2}
\end{align}
$$
That is,
$$
\frac1{n\log(n)}\ge\log(\log(n+1))-\log(\log(n))\tag{3}
$$
Summing $(3)$, and noticing that the sum on the right telescopes, we see that
$$
\begin{align}
\sum_{n=2}^m\frac1{n\log_2(n)}
&=\log(2)\sum_{n=2}^m\frac1{n\log(n)}\\
&\ge\log(2)[\log(\log(m+1))-\log(\log(2))]\tag{4}
\end{align}
$$
Since the right side of $(4)$ grows without bound, the series on the left side diverges.

Asymptotic Behavior
First of all,
$$
\sum_{k=2}^n\frac1{k\log_2(k)}
=\log(2)\sum_{k=2}^n\frac1{k\log(k)}\tag{5}
$$
Using The Euler Maclaurin Sum Formula, we get
$$
\hspace{-1cm}\small\sum_{k=2}^n\frac1{k\log(k)}
=\log(\log(n))+C+\frac1{2n\log(n)}-\frac1{12n^2\log(n)}-\frac1{12n^2\log(n)^2}+O\left(\frac1{n^4\log(n)}\right)\tag{6}
$$
where
$$
C=0.7946786454528994022038979620651495140650\tag{7}
$$
In any case, $(6)$ means that $(5)$ diverges.
