How to apply Parseval's equation and checking to see if we have convergence in $L^2$ 
On a previous assignment, you 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 $L^2$.
Attempted solution 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*}
Attempted solution c.) - Yes, the series converges uniformly by the Weirerstrass $M$-test, see the solution to part a.).
First off I want to know if these solutions are sufficient. Second, I want to know how to apply Pareseval's equation to find the value of $$\sum_{n=1}^{infty}\frac{1}{(2n+1)^6}$$

Parseval's equation,$$\sum_{n=1}^{\infty}|\hat{f}_n|^2 = \|f\|^2$$ We have 
$$\sum_{n=1}^{\infty}\left(\frac{8l^2}{\pi^3}\frac{1}{n^3}\right)^2 = \sum_{k=0}^{\infty}\left(\frac{8l^2}{\pi^3}\frac{1}{(2k+1)^3}\right)^2 = \frac{2}{l}\int_{0}^{l}(x(l-x))^2dx = \frac{l^4}{15}$$
Therefore, $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^6} = \frac{\pi^6}{64}\frac{1}{15} = \frac{\pi^6}{960}$$
