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!
 A: 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.
A: 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
$$\sum_{n=0}^\infty a_n = \int_0^1\frac{\sin(\pi x)}{(1-x)^2}dx=\infty$$
A: Remember that you can view sequences as functions $\mathbb{Z}_{\geq0}\rightarrow\mathbb{R}$; in this sense, for a sequence $a:\mathbb{Z}_{\geq0}\rightarrow\mathbb{R}$, the series $\sum_{n=0}^{\infty}a(n)$ is precisely $\int_{\mathbb{Z}_{\geq0}}a\,d\nu$, where $\nu$ is the counting measure on $\mathbb{Z}_{\geq0}$.
This, we can consider
$$
\sum_{n=0}^{\infty}\int_0^1\frac{x^n}{1-x}\sin(\pi x)\,dx=\int_{\mathbb{Z}_{\geq0}}\int_{[0,1]}\frac{x^n}{1-x}\sin(\pi x)\,d\mu(x)\,d\nu(n),
$$
where $\mu$ is Lebesgue measure on $[0,1]$.
Now, the integrand here is non-negative; so, by Tonelli's theorem, we can switch the order of integration. Thus
$$
\int_{\mathbb{Z}_{\geq0}}\int_{[0,1]}\frac{x^n}{1-x}\sin(\pi x)\,d\mu(x)\,d\nu(n)=\int_{[0,1]}\int_{\mathbb{Z}_{\geq0}}\frac{x^n}{1-x}\sin(\pi x)\,d\nu(n)\,d\mu(x).
$$
However, we know that
$$
\int_{\mathbb{Z}_{\geq0}}\frac{x^n}{1-x}\,d\nu(n)=\sum_{n=0}^{\infty}\frac{x^n}{1-x}\sin(\pi x)=\frac{\sin(\pi x)}{(1-x)^2}.
$$
So, at this point, we have reduced the problem to trying to determine whether or not the following integral converges:
$$
\int_0^1\frac{\sin(\pi x)}{(1-x)^2}\,dx.
$$
You can take it from there!
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{a_{n} = \int_{0}^{1}{x^{n} \over 1 - x}\,\sin\pars{\pi x}\,\dd x:\ {\large ?}}$.

\begin{align}
a_{n} &= \int_{0}^{1}\pars{1 - x}^{n}\,{\sin\pars{\pi x} \over x}\,\dd x
\\[3mm]
\verts{a_{n}} &< \pi\int_{0}^{1}\pars{1 - x}^{n}\,\dd x
= \left.\pi\,{\pars{1 - x}^{n + 1} \over -n - 1}\right\vert_{x = 0}^{x = 1}
= {\pi \over n + 1}
\end{align}

$$
\color{#0000ff}{\large\lim_{n \to \infty}a_{n} = 0}
$$
