2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ There is always more than one way to solve a problem. But that said, derivatives are the go-to tool for finding extrema, and since this series is differentiable except at certain isolated points, pretty much anyone would look there first. $\endgroup$ – Paul Sinclair Mar 17 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.