On the uniform convergence of the series of the following functions prove that $$\sum_{n=1}^{\infty}\int_{0}^{x}t^n\sin\pi tdt$$Uniform convergence on$[0,1]$.
I try to use Weierstrass discriminant
$$\left|\int_{0}^{x}t^n\sin\pi tdt\right|\leqslant\int_{0}^{x}t^ndt=\frac{x^{n+1}}{n+1}$$
Uniform convergence when $x < 1$.I don't know if this is wrong,I want to know how to judge the convergence and divergence at 1.Thanks.
 A: Step 1.
We first notice that
\begin{align}
f_n(x) & = \sum_{i=1}^{n}\int_{0}^{x}t^i\sin\pi tdt\\
       & = \int_{0}^{x} \sin\pi t \left( \sum_{i=1}^{n}t^i \right)dt.
\end{align}
By Dominated convergence theorem, we get
\begin{align}
\lim_{n\rightarrow \infty} f_n(x)
& = \int_{0}^{x} \sin(\pi t) \left( \sum_{i=1}^{\infty}t^i \right)dt\\
& =  \int_{0}^{x} \sin(\pi t) \left( \frac{1}{1-t}-1\right)dt=: f(x).\\
\end{align}
Step 2. We now prove that $f_n$ converges uniformly to $f$, i.e.
$$\sup_{x\in [0,1]} |f(x)-f_n(x)|\xrightarrow{n\rightarrow \infty} 0.$$
Indeed,
\begin{align}
|f(x)-f_n(x)|
& = |\int_{0}^{x} \sin\pi t \left( \sum_{i=n+1}^{\infty}t^i \right)dt|\\
& \leq \int_{0}^{x} t^{n+1} \left( \sum_{i=0}^{\infty}t^i \right)dt.\\
& =  \int_{0}^{1} \frac{t^{n+1}}{1-t} dt.
\end{align}
Thus,
$$\sup_{x\in [0,1]} |f(x)-f_n(x)|\leq \int_{0}^{1} \frac{t^{n+1}}{1-t} dt
\xrightarrow{n\rightarrow \infty} \int_{0}^{1} \lim_{n\rightarrow \infty} \frac{t^{n+1}}{1-t}  dt=0.$$
A: I got it,let $$u_{n}(x)=\int_{0}^{x}t^n\sin\pi tdt$$ be aware$$0\leqslant u_{n}(x)\leqslant u_n(1)=\int_{0}^{1}t^n\sin\pi tdt=\left|\frac{\pi}{n+1}\int_{0}^{1}t^{n+1}\cos\pi tdt\right|<\frac{\pi}{(n+1)(n+2)}.$$according to Weierstrass discriminant,$$\sum_{n=1}^{\infty}\int_{0}^{x}t^n\sin\pi tdt$$Uniform convergence on$[0,1]$.
