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?
 A: $$\sum_{n=1}^\infty\int_0^{\frac{1}{n}}\frac{\sqrt x}{1+x^2}dx<\sum_{n=1}^\infty\int_0^{\frac{1}{n}}\frac{\sqrt x}{1}dx=\sum_{n=1}^\infty\frac{2}{3n^{3/2}}=\frac{2}{3}\zeta\left(\frac{3}{2}\right)$$
A: 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.
A: Since $\sqrt{x}$ is increasing and $1/(1+x^2)$ is decreasing, we have
$$ \int_0^{1/n} \frac{\sqrt{x}}{1+x^2} \, dx < \frac{1}{\sqrt{n}} \int_0^{1/n}\frac{dx}{1+x^2} < \frac{1}{\sqrt{n}} \int_0^{1/n} dx = \frac{1}{n^{3/2}}.  $$
 And $\sum_n 1/n^{3/2}$ converges.
A: I think you tried to estimate the wrong part of the fraction. If you think about it, $\sqrt{x}$ is much different from 1 and from $x$ in a close neighborhood of 0. However, $1+x^2$ is very close to 1 there. 
So simply estimate by $\int\limits_{0}^{1/n} \sqrt{x}= \frac{2}{3}n^{-3/2}$, whose sum is convergent. 
