Showing the sine series on the interval $[0,l]$ converges in $L^2$ 
We found the sine series on the interval $[0,l]$:
  $$x(l-x) = \frac{8l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n \pi x}{l}\right)$$
  Note that the sum includes odd terms only.
a.) Show that this series converges in $L^2$. Then, apply Parseval's equation to find the value of $\sum_{n=0}^{\infty}\frac{1}{(2n +1)^6}$.
b.) Show that this series converges pointwise. Then, plug in $x = \frac{l}{2}$ to find the value of $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}$.
c.) Does the series converge uniformly? Why or why not?

Attempted Solution:
a.) We have 
$$\Bigg| \frac{1}{n^3}\sin\left(\frac{n \pi x}{l}\right)\Bigg| \leq \frac{1}{n^3} $$
Now take $M_n = \frac{1}{n^3}$ since $\sum_{n=1}^{\infty}\frac{1}{n^3}$ is convergent then by the Weirerstrass $M$-test the original series is uniformly convergent and thus convergent in $L^2$.
(Just don't know how to apply Parseval's equation here)
b.) Since we have shown that the series is uniformly convergent then we automatically have pointwise convergence since uniform convergence implies pointwise convergence. \
\noindent
Now, letting $x = \frac{l}{2}$, we have 
\begin{align*}
&\frac{l}{2}\left(l - \frac{l}{2} \right) = \frac{8 l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n\pi (l/2)}{l} \right)\\
&\Rightarrow \frac{l^2}{4} = \frac{8 l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}\sin\left(\frac{n\pi}{2}\right)
\end{align*}
Note that 
$$\sin\left(\frac{n\pi}{2}\right) = \begin{cases}
0 \ \ &\text{if} \ \ n \ \ \text{even}, n = 2m, m = 0,1,2,\ldots\\
(-1)^n \ \ &\text{if} \ \ n \ \ \text{odd}, n = 2m + 1, m = 0,1,2,\ldots
\end{cases}$$
Therefore we have 
\begin{align*}
&\frac{l^2}{4} = \frac{8 l^2}{\pi^3}\sum_{n=1}^{\infty}\frac{1}{n^3}(-1)^n\\
&\Rightarrow \sum_{n=1}^{\infty}\frac{(-1)^n}{n^3} = \frac{\pi^3}{32}
\end{align*}
c.) Yes, the series converges uniformly by the Weirerstrass $M$-test, see the solution to part a.).
Please let me know if this is a sufficient solution for this problem, any comments are appreciated.
 A: It seems that the questions are structured in a way that increases in difficulty. If you prove its uniform convergence, $(c)$, like in the comments, then basically $(a)$ and $(b)$ follows immediately. 
So if you want a bottom-up approach, then it could go something like this. 
a) Note that $|a_n|^2\leq\dfrac{1}{n^6}$ and the harmonic series is convergent. By comparision test, $\sum |a_n|^2$ is convergent. 
Now Parseval's equation is simply $$\dfrac{1}{2l}\int_0^l|f(x)|^2dx = \sum_{n=-\infty}^{\infty}|c_n|^2,$$
where $f$ is your function and $c_n$ are the Fourier Coefficients. This will be very easy because you are already given the coefficients $c_n$ in your equation.
b) Pointwise convergent is again immediate again, by the comparison test as $\sum \dfrac{1}{n^3}$ is convergent. In order to evaluate the given sum, you just need to prove $\sin(\dfrac{n\pi}{2}) = (-1)^{\frac{n-1}{2}}$, for odd $n$ and zero otherwise.   
A: Lemma 1. $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ is the Fourier series of a sawtooth wave, a $2\pi$-periodic function that equals $\frac{\pi-x}{2}$ on the interval $(0,2\pi)$. The convergence in $L^2$ simply follows from computing $\int_{0}^{2\pi}\frac{\pi-x}{2}\sin(nx)\,dx$ by integration by parts. For any $x\in\mathbb{R}\setminus\pi\mathbb{Z}$ we also have pointwise convergence by Dirichlet's test, since the partial sums of $\sin(nx)$ are bounded. For the same reason we have uniform convergence over any interval of the form $[a\pi+\varepsilon,(a+1)\pi-\varepsilon]$ for $a\in\mathbb{Z}$ and $\varepsilon>0$. The convergence over $\mathbb{R}\setminus\pi\mathbb{Z}$ is not uniform due to Gibbs phenomenon.
What happens by integrating $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ twice?
It happens that $\sum_{n\geq 1}\frac{\sin(nx)}{n^3}$ turns out to be convergent, both uniformly and in $L^2$, to a $2\pi$-periodic function that is piecewise-polynomial and piecewise-cubic. Namely:
Lemma 2.
$$ \forall x\in[0,2\pi],\qquad \sum_{n\geq 1}\frac{\sin(nx)}{n^3} = \text{Im }\text{Li}_3(e^{ix}) = \frac{x(x-\pi)(x-2\pi)}{12}. $$
By evaluating both sides at $x=\frac{\pi}{2}$ we get
$$ \sum_{m\geq 0}\frac{(-1)^m}{(2m+1)^3} = \frac{\pi^3}{32} $$
that is also proved here with a different approach. By Parseval's theorem
$$ \zeta(6)=\sum_{n\geq 1}\frac{1}{n^6}=\frac{1}{\pi}\int_{0}^{2\pi}\left(\frac{x(x-\pi)(x-2\pi)}{12}\right)^2\,dx = \frac{\pi^6}{945} $$
hence
$$ \sum_{n\geq 0}\frac{1}{(2n+1)^6} = \zeta(6)-\frac{1}{64}\zeta(6) = \frac{\pi^6}{960}.$$
