Finding the uniform convergence of a Fourier series Verify that the fourier series converge uniformly on the interval ${\pi\leq x\leq \pi}$. Also state why this series is differentable in the interval ${\pi\leq x\leq \pi}$, except at the point $x=0$ and describe graphically the function that is represented by the differentiated series for all $x$
$\frac{1}{\pi}+0.5\sin x-\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }$
What i tried
Using the Weierstrass M test i got $$|\frac{1}{\pi}+0.5\sin x-\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|\leq |\frac{1}{\pi}+0.5\sin x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|$$
$$|\frac{1}{\pi}+0.5\sin x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }| \leq |\frac{1}{\pi}+0.5\ x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }|$$
$$|\frac{1}{\pi}+0.5\ x+\frac{2}{\pi}\sum_{n=1}^\infty \dfrac{\cos 2nx}{4n^{2}-1 }| \leq|\frac{1}{\pi}+0.5\ x+\frac{2x}{\pi}\sum_{n=1}^\infty \dfrac{\ 2n}{4n^{2}-1 }|$$
And since the term 
$$\sum_{n=1}^\infty \dfrac{\ 2n}{4n^{2}-1 }$$ converges then by the Weierstrass M test the above fourier seriesM test converges uniformly.
Differentating the fourier series term by term i got
$$0.5\cos x+\frac{4n}{\pi}\sum_{n=1}^\infty \dfrac{\sin 2nx}{4n^{2}-1 }$$
I suppose that it is differentiable on the given interval because it is continous on that interval except at $x=0$ but i cant see why is this so and also how to describe  the graphically the function that is represented by the differentiated series for all $x$. Could anyone please explain this to me. Thanks
 A: The work in the OP has some flaws.  Note that we have for all $x$, $\left|\frac{\cos(2nx)}{4n^2-1}\right| \le \frac1{4n^2-1}$, for each $n$.  Inasmuch as 
$$\begin{align}
\sum_{n=1}^\infty \frac{1}{4n^2-1}&<\infty
\end{align}$$
the Weierstrass M-Test guarantees that the series $\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$ converges uniformly for all $x\in [-\pi,\pi]$.

To analyze whether the series is differentiable, we examine the series of term-by-term derivatives $D(x)$ as given by
$$D(x)=-2\sum_{n=1}^\infty \frac{n\sin(2nx)}{4n^2-1} \tag 1$$
Note that $\sum_{n=1}^N \sin(2nx)=\csc(x)\sin(Nx)\sin((N+1)x)$ is bounded by $|\csc(x)|$ for $x\ne 0,\pi,-\pi$.  Furthermore, $\frac{n}{4n^2-1}$ monotonically decreases to $0$ as $n\to \infty$.  
Therefore, for any $\delta >0$ and $x\in [-\pi+\delta,-\delta]$ or $x\in [\delta,\pi-\delta]$, Dirichlet's Test guarantees that  the series in $(1)$ for $D(x)$ converges uniformly and inasmuch as the original series $\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$ also converges on $[-\pi,\pi]$ (actually, we only need that it converge at a single point), we find that
$$D(x)=\frac{d}{dx}\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2-1}$$
for all $x\ne 0,\pi,-\pi$.

To examine the derivative from the right (left) of $\sum_{n=1}^\infty \frac{\cos(2nx)}{4n^2- 1}$ at $x=-\pi$ ($x=\pi$), we note that 
$$\begin{align}
\lim_{h\to0^{\pm}}\sum_{n=1}^\infty \frac{\cos(2n(\mp \pi+h))-\cos(2n(\mp\pi))}{h(4n^2- 1)}&=-2\lim_{h\to0^{\pm}}\sum_{n=1}^\infty \frac{\sin^2(nh)}{h(4n^2-1)}\\\\
&\overbrace{=}^{\text{LHR}}\underbrace{-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\frac{4n\sin(2nh)}{4n^2-1}}_{=\lim_{x\to \mp\pi}D(x)}\\\\
&=-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\left(\frac{\sin(2nh)}{n}+\frac{\sin(2nh)}{n(4n^2-1)}\right)\\\\
&=-\frac12\lim_{h\to0^{\pm}}\sum_{n=1}^\infty\frac{\sin(2nh)}{n}\\\\
&=\mp \frac\pi4
\end{align}$$ 
It is importatn to observe that $D(\pm\pi)=0\ne \pm\frac\pi4$. This is not inconsistent since $D(x)$ is not the representation of the derivative (from the left or right)) at $x=\pi$ or $x=-\pi$.  However, $D(x)$ does have the appropriate limits as $x\to \pm\pi$.
