$I_n=\frac{1}{\pi}-\frac{n(n-1)}{\pi^2}I_{n-2}$ I have to calculate $I_n=\int_0^1 x^n \sin{(\pi x)} dx$.
By double partial integration I got $I_n=\frac{1}{\pi}-\frac{n(n-1)}{\pi^2}I_{n-2}$ and $I_0=\frac{2}{\pi}$, $I_1=\frac{1}{\pi}$.
How to solve this recurrence, in order to find $\lim_{n\to\infty} I_n$ and to deduce convergence of $\sum_{n=1}^\infty I_n$?
Any help is welcome. Thanks in advance.
$$(\forall n\in \Bbb N)\;\;\big|I_n\big|=\left|\int_0^1x^n\sin(\pi x)dx\right|$$
$$\le \int_0^1\big|x^n\sin(\pi x)\big|dx$$
$$\le \int_0^1x^ndx=\frac{1}{n+1}$$
$$\implies \lim_{n\to+\infty}I_n=0$$
The recursive relation you got by double partial integration, can also be written as
$$\boxed{I_{n+2}\pi^2=\color{red}{\pi}-(n+2)(n+1)I_n.}$$
$$\lim_{n\to+\infty}I_{n+2}=0\implies$$
$$\implies \lim_{n\to+\infty}(n+2)(n+1)I_n=\color{red}{\pi}$$
$$\implies I_n\sim \frac{\pi}{(n+2)(n+1)}\sim \frac{\pi}{n^2}$$
$$\implies \sum I_n \text{ converges}$$
$\;\sum\limits_{n=1}^\infty I_n=\text{Si}(\pi)-\dfrac2\pi\;$
and as far as I know there is not a simplier  way to write the sum of your series.