checking the convergence of this series I am trying to check the convergence of this series
$$\sum_{n=2}^{n=\infty}\frac{ln(n)}{n\sqrt{1+n}}$$
I tried divergence test , it gives me zero ( it fails).
I tried limit comparison test with 
$$b_n=\sum\frac{1}{n^2}$$
but it fails (the limit gives me infinity , and bn is convergent)
I tried limit comparison with $$b_n=\sum\frac{1}{n}$$
But it fails ( the limit is zero , and bn is divergent).
I tried comparison test with (ln(n)/n) but failed also .
How can I check the convergence of this series?!
 A: What about using Cauchy's Condensation Test?:
$$\sum_{n=1}^\infty 2^na_{2^n}=\sum_{n=1}^\infty\frac{2^n\log2^n}{2^n\sqrt{1+2^n}}=\sum_{n=1}^\infty\frac{n\log2}{\sqrt{1+2^n}}$$
But the last series converges since, for example
$$\frac1{\sqrt2}\xleftarrow[n\leftarrow\infty]{}\frac{\sqrt[n]{n\log2}}{\sqrt2\,\sqrt[n]2}=\sqrt[n]{\frac{n\log2}{\sqrt{2\cdot2^n}}}\le\color{red}{\sqrt[n]{\frac{n\log2}{\sqrt{1+2^n}}}}\le\sqrt[n]\frac{n\log2}{\sqrt{2^n}}=\frac{\sqrt[n]{n\log2}}{\sqrt2}\xrightarrow[n\to\infty]{}\frac1{\sqrt2}$$
A: Let $u_n $ be the general term.
we have $$u_n\sim \frac {\ln (n)}{n^\frac 32} $$
and
$$\lim_{n\to +\infty }n^\frac 75 u_n=0,$$
because $\frac 75 <\frac 32.$
thus for great enough $n $,
$$n^\frac 75 u_n <1$$
or
$$0 <u_n <\frac {1}{n^\frac 75} $$
this proves the convergence of $\sum u_n$.
A: In this case the asymptotic exponent for $a_n$ is in between 1 "divergent case" and 2 "convergent case" so the series is convergent to show this you have to select an intermediate exponent for the comparison test.
A: Creative telescoping gives another way. We have
$$ \frac{\log n}{n\sqrt{n+1}}\leq 2\left[\frac{3+\log n}{\sqrt{n}}-\frac{3+\log(n+1)}{\sqrt{n+1}}\right]$$
for any $n\geq 2$, hence
$$ \sum_{n=2}^{N}\frac{\log n}{n\sqrt{n+1}}\leq 2\left[\frac{3+\log 2}{\sqrt{2}}-\frac{3+\log(N+1)}{\sqrt{N+1}}\right]\leq (3+\log 2)\sqrt{2} $$
proves that the LHS is a convergent series.
