convergence of $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx$ Test convergence of $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx$
Attempt: Since, $\lim_{n \rightarrow \infty} \dfrac {1}{n} \in (0,1) \implies \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx \geq \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac { x}{1+x^2} dx$
If $I = \int_0^{\frac {1}{n}} \dfrac { x}{1+x^2} dx = \dfrac {1}{2} \ln (1+ \dfrac {1}{n^2})$
Now, for $\ln (1+\dfrac {1}{n^2}) <(1+\dfrac {1}{n^2})$
Since, $\sum_{n=1}^\infty (1+\dfrac {1}{n^2})$ is convergent $\implies \sum_{n=1}^\infty   \ln (1+\dfrac {1}{n^2})$ is convergent as well.
But, I don't think this result is of any use as $\sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac {\sqrt x}{1+x^2} dx \geq \sum_{n=1}^\infty \int_0^{\frac {1}{n}} \dfrac { x}{1+x^2} dx$
Please guide me on moving from here.
Thank you very much for your help.

First note that the denominator is there just to confuse you. For sufficiently small $x$,
$$1\ge\frac1{1+x^2}\ge\frac12,$$
so we can disregard that altogether. Now
$$\sum_{n=1}^\infty\int_0^{1/n}\sqrt x\,dx=\frac23\sum_{n=1}^\infty \frac 1{n^{3/2}},$$
which converges.
