Investigate the convergence of $\sum a_n$ where $a_n = \int_0^1 \frac{x^n}{1-x}\sin(\pi x) \,dx$ Investigate the convergence of $\sum a_n$ where $a_n = \displaystyle\int_0^1 \dfrac{x^n}{1-x}\sin(\pi x) \,dx$.
We have thought about using the dominated convergence theorem to find $\lim a_n$, but that would result in something like $\lim a_n = \lim \displaystyle\int_0^1 \dfrac{x^n}{1-x}\sin(\pi x) \,dx = \displaystyle\int_0^1 \lim \dfrac{x^n}{1-x}\sin(\pi x) \,dx = \displaystyle\int_0^1 0 \,dx = 0$ which makes the convergence of $\sum a_n$ inconclusive.
Any tips on how to proceed? Thanks!

Using $\sin(\pi x) \geqslant \pi x(1-x)$ for $0\leqslant x \leqslant 1$ 
$$
   a_n = \int_0^1 \frac{x^n}{1-x} \sin(\pi x) \mathrm{d}x > \int_0^1 \pi x^{n+1} \mathrm{d}x = \frac{\pi}{n+2}
$$
it follows that $\sum_{n=0}^\infty a_n$ diverges.
