Convergence of the series $\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$. 
Check whether the series
  $$\sum_{n=1}^{\infty}\int_0^{\frac{1}{n}}\frac{\sqrt{x}}{1+x^2}\ dx$$
  is convergent.

I tried to sandwich the function by $\dfrac{1}{1+x^2}$ and $\dfrac{x}{1+x^2}$ , but this did not help at all.
Any other way of approaching?

When $x\in [0,1], \frac {\sqrt x}{2} \le\frac {\sqrt x}{1+x^2} \le \sqrt x$
Which means that $\frac 12\int_0^\frac1n \sqrt x\ dx \le \int_0^\frac1n \frac {\sqrt x}{1+x^2} \ dx \le \int_0^\frac1n \sqrt x \ dx$
if $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \sqrt x \ dx $ converges then  $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \frac{\sqrt x}{1+x^2} \ dx $ converges.
and if $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \sqrt x \ dx $ diverges then $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \frac 12\sqrt x \ dx $ diverges and   $\sum_\limits{n=1}^{\infty} \int_0^\frac1n \frac{\sqrt x}{1+x^2} \ dx $ diverges.
