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!

Though Sasha's solution is perfect, I'd like to suggest another approach (suitable to the [measure-theory] tag):
We note that
$$S_N = \sum_{n=0}^N a_n = \int_0^1 \frac{1-x^{N+1}}{1-x}\frac{\sin(\pi x)}{1-x}dx$$
Now, $f_N := \frac{1-x^{N+1}}{1-x}\frac{\sin(\pi x)}{1-x}\to\frac{\sin(\pi x)}{(1-x)^2}$ pointwise and monotonically, so by the monotone convergence theorem we have
$$\sum_{n=0}^\infty a_n = \int_0^1\frac{\sin(\pi x)}{(1-x)^2}dx=\infty$$
