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 $0\leqslant x \leqslant 1$ 
$$
   a_n = \int_0^1 \frac{x^n}{1-x} \sin(\pi x) \mathrm{d}x &gt; \int_0^1 \pi x^{n+1} \mathrm{d}x = \frac{\pi}{n+2}
$$
it follows that $\sum_{n=0}^\infty a_n$ diverges.
$\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{a_{n} = \int_{0}^{1}{x^{n} \over 1 - x}\,\sin\pars{\pi x}\,\dd x:\ {\large ?}}$.


\begin{align}
a_{n} &amp;= \int_{0}^{1}\pars{1 - x}^{n}\,{\sin\pars{\pi x} \over x}\,\dd x
\\[3mm]
\verts{a_{n}} &amp;&lt; \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}


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