It's kinda weird summation

In one sense it did. I knew that Fourier series like this exist, and I initially solved the problem by looking it up in Gradshteyn and Rhyzik.

We could work out what $f(x)$ had to be with a little thought, though. Suppose that $f(x)$ is an odd polynomial in $x$ . The Fourier sine series of $f(x)$ is $f(x) \; = \; \sum_{k=1}^\infty a_k \sin kx$ where $a_k \; = \; \frac{1}{\pi} \int_{-\pi}^\pi f(x)\,\sin kx\,dx \qquad \qquad k \ge 1 \;.$ Integrating by parts gives $a_k \; = \; \Big[ -\frac{1}{\pi k}f(x) \cos kx \Big]_{-\pi}^\pi + \frac{1}{k\pi}\int_{-\pi}^\pi f'(x) \cos kx\,dx \;;$ Since we want $a_k \,=\, (-1)^{k+1} k^{-3}$ , we had better have $f(\pi) = f(-\pi) =0$ to make the cross term disappear. Thus $\pi^2 - x^2$ must be a factor of $f(x)$ . Integrating by parts two more times gives $\begin{array}{rcl} a_k & = & \displaystyle \frac{1}{k\pi}\int_{-\pi}^\pi f'(x)\cos kx\,dx \; = \; \Big[\frac{1}{k^2\pi} f'(x)\sin kx \Big]_{-\pi}^\pi - \frac{1}{k^2\pi}\int_{-\pi}^\pi f''(x)\, \sin kx\,dx \\ & = & \displaystyle-\frac{1}{k^2\pi}\int_{-\pi}^\pi f''(x)\,\sin kx\,dx \; = \; \Big[\frac{1}{k^3\pi} f''(x)\cos kx\Big]_{-\pi}^\pi - \frac{1}{k^3\pi}\int_{-\pi}^\pi f'''(x)\cos kx\,dx \end{array}$
If we try having $f(x)$ cubic then (since $f(x)$ is odd), we need $f(x) = \alpha x(\pi^2 - x^2)$ , so that $f''(x) = -6\alpha x$ and $f'''(x) = -6\alpha$ , and hence $a_k \; = \; \Big[ - \frac{6\alpha}{k^3 \pi}x \cos kx \Big]_{-\pi}^\pi + \frac{6\alpha}{k^3\pi}\int_{-\pi}^\pi \cos kx\,dx \; = \; (-1)^{k+1} \frac{12\alpha}{k^3}$ which tells us that $\alpha = \tfrac{1}{12}$ , as desired.

Its (-Li3(-e^i) + Li3(-e^-i))/2i Where i represents iota and three is subscript of Li

@Pi Han Goh Sir, to determine the initial function, we can also use an approach similar to the one given in this solution.........