I'm trying to prove that for $x\in[0,\pi]$
$$ \frac{\sin (x)}{3} - \frac{\sin (3x)}{1 \cdot 3 \cdot 5} - \frac{\sin (5x)}{3 \cdot 5 \cdot 7} - \frac{\sin (7x)}{5 \cdot 7 \cdot 9} \cdots = \frac{\pi}{8} \sin^2 (x) $$
and determine if the series converges uniformly, pointwise, and/or in $L^2$.
I've identified that we can write it as
$$ \frac{\sin(x)}{3} + \sum_{n=0}^\infty - \frac{\sin((2n+3)x)}{(2n+1)(2n+3)(2n+5)} $$
which looks an awful lot like the Fourier series of $ f(x)=\frac{\pi}{8} \sin^2(x). $
Indeed, if we pretend $f(x)$ is odd (even though it's even), and runs on the interval $[-\pi,\pi]$, we get the right coefficients. We have
$$ f(x)=\sum_{k=1}^\infty b_k \sin(kx) $$
where
$$ b_k = \frac{2}{\pi} \int_0^\pi f(x) \sin(kx) dx. $$
Immediately, we get that $b_k=0$ for even $k$, and we find
$$ b_k = \frac{2 (\pi -\pi \cos (\pi k))}{\pi \left(16 k-4 k^3\right)} $$
which gives the same expansion as our initial sum.
I feel like there should be a way of converting this to the real Fourier series expansion to show convergence, however, I'm unable to figure out how.
Furthermore, I've tried to look at uniform convergence from the definition, but I haven't been able to make much progress as I don't know how to deal with the $\sin$ in the sum.
$$ \Bigg| \frac{\sin(x)}{3} + \sum_{n=0}^\infty - \frac{\sin((2n+3)x)}{(2n+1)(2n+3)(2n+5)} - \frac{\pi}{8} \sin^2 (x) \Bigg| < \epsilon. $$
Finally, I've looked at the problem numerically, and have been able to show that the series likely converges uniformly, and therefore also pointwise.
How would I be able to continue my analysis?
Without Fourier analysis. $$\sum_{n=0}^\infty \frac{\exp(i(2n+3)x)}{(2n+1)(2n+3)(2n+5)}=\frac{1}{12} \left(4 i \sin (x)+\cos (x)-6 \sin ^2(x) \tanh ^{-1}\left(e^{i x}\right)\right) $$ $$\sum_{n=0}^\infty \frac{\sin((2n+3)x)}{(2n+1)(2n+3)(2n+5)}=\frac{1}{12} \sin (x) \left(4-3 i \sin (x) \left(\tanh ^{-1}\left(e^{-i x}\right)-\tanh ^{-1}\left(e^{i x}\right)\right)\right)$$
Now, using $$\tanh ^{-1}(z)=\frac{1}{2} \log \left(\frac{1+z}{1-z}\right)$$ $$\tanh ^{-1}\left(e^{-i x}\right)=\frac{1}{2} \log \left(-i \cot \left(\frac{x}{2}\right)\right)\qquad \text{and} \qquad \tanh ^{-1}\left(e^{i x}\right)=\frac{1}{2} \log \left(i \cot \left(\frac{x}{2}\right)\right)$$ $$\tanh ^{-1}\left(e^{-i x}\right)-\tanh ^{-1}\left(e^{i x}\right)=\frac 12 \log(-1)$$
Just continue using the appropriate branch for $\log (-1)$.