The question at hand is to find the maximum value of:
$$ \lim_{n \to \infty}\sum_{k=1}^{n} \frac{\sin{kx}}{k^{3}} = \frac{\sin{x}}{1^{3}} + \frac{\sin{2x}}{2^{3}} + \frac{\sin{3x}}{3^{3}} ........... \infty $$
My approach was as follows:
Let $f(x) = \sum_{k=1}^{\infty} \frac{\sin({kx})}{k^3}$
Then $f'(x) = \sum_{k=1}^{\infty} \frac{\cos({kx})}{k^2}$
And $ f''(x) = \sum_{k=1}^{\infty} \frac{(-1)\sin({kx})}{k}$
Now, it is known that $ \sum_{k=1}^{\infty} \frac{\sin({kx})}{k} $ is the Fourier series expansion of the function $ f(x) = \frac{\pi - x}{2}$ for $ x \in (0, 2\pi)$
So $ f''(x) = \frac{x - \pi}{2}$
Hence after integrating once, we get:
$$ f'(x) = \frac{1}{4}x^2 - \frac{\pi}{2}x + C$$
Here $C = f'(0) = \sum_{k=1}^{\infty} \frac{1}{k^2} = \zeta(2) = \frac{\pi^2}{6} \implies f'(x) = \frac{1}{4}x^2 - \frac{\pi}{2}x + \frac{\pi^2}{6} $
Now integrating $f'(x)$ once more, we get:
$$ f(x) = \frac{1}{12}x^{3} - \frac{\pi}{4}x^{2} + \frac{\pi^2}{6}x + C$$
Since $f(0) = 0 \implies C = 0$
And hence we end up with $$ f(x) = \frac{1}{12}x^{3} - \frac{\pi}{4}x^{2} + \frac{\pi^2}{6} x$$
At this point, I now have a function of $x$ which I can differentiate to find out the maximum value using the second derivative test.
My question would be is there any other way to approach this problem? It would be awesome to see how community decides to approach this problem.
Furthermore, this method requires the knowledge of Fourier series expansions and I'd like to see any other methods not involving Fourier expansions.