Convergence of $\sum_{n=2}^\infty \frac{1}{\sqrt{n\log n}}$ I'm having difficulties examining convergence of the series 
$$\sum_{n=2}^\infty \frac{1}{\sqrt{n\log n}}.$$
This root/ratio test is out of the question here. I tried to manipulate the denominator using properties of the logarithm but it didn't help. 
Note: I'm not allowed to use the integral test (I also think that integration would be quiet challenging here). 
 A: Since $\log n <\sqrt{n}$ for $n\geq2$, you have $$\frac{1}{\sqrt{n\log n}} >
\frac{1}{\sqrt{n}\sqrt{n}} = \frac{1}{n}.$$  This sum of the last guy diverges.
A: $\lim_n{{\log(n)}\over {n^{1/4}}}={1\over 4}\lim_n{{\log(n^{1/4})}\over n^{1/4}}={1\over 4}\lim_n {{\log(n)}\over n}=0$, there exists $N$ such that $n>N$ implies that $\log(n)<n^{1/4}$, we deduce that ${1\over\sqrt{n\log(n)}}>{1\over{n^{3/4}}}$. Compare with the sequence $\sum_n{1\over n^{3/4}}$.
A: note that $\log { n } <\sqrt { n } ,n\ge 2$ so $$\sum _{ n=2 }^{ \infty  } \frac { 1 }{ \sqrt { n\log  n }  } >\sum _{ n=2 }^{ \infty  } \frac { 1 }{ n } $$ means series  diverges
A: Observe that $$ \sum_{i=2}^\infty\frac{1}{n^2}\le \sum_{i=2}^\infty\frac{1}{n\log n}\implies \sum_{i=2}^\infty\frac{1}{n}\le \sum_{i=2}^\infty\frac{1}{\sqrt{n\log n}} $$
Since, the harmonic series diverges which means the given series diverges.
A: You can use the cauchy condensation test which works great with logs. Let $a_n$ be the general term of your series. Then your series converges iff the series $\sum 2^n{a_{2^n}}$ converges.  But $$ 2^n{a_{2^n}}=\frac{2^n}{\sqrt{2^n\log(2^n)}}=\frac{2^{n/2}}{\sqrt n\sqrt{\log2}}\longrightarrow \infty$$ so that your series diverges.
