17th March

Let's generalise the sum . The sum is in a form of $\displaystyle\sum_{n=1}^∞ \dfrac{\sin(nθ)}{n^3}$

Generally , $\displaystyle\sum_{n=1}^∞ \dfrac{\sin(n\theta)}{n}$ $= \dfrac{1}{2i} \sum_{n=1}^∞ \dfrac{e^{i n\theta}-e^{-i n\theta}}{n}= -\dfrac{1}{2i}(\log(1-e^{i\theta})-\log(1-e^{-i\theta}))$ $= \dfrac{1}{2i} \log(-e^{-i\theta})= \dfrac{\log(e^{iπ})-\log(e^{i\theta})}{2i} = \dfrac{π-\theta}{2}$ Now integrate both sides from $0$ to $\theta$ $\sum_{n=1}^∞ \int_0^{\theta} \dfrac{\sin(n\theta)}{n} d\theta= \int_0^{\theta} \dfrac{π-\theta}{2}d\theta$ $\sum_{n=1}^∞ \dfrac{\cos(n\theta)}{n^2} = \dfrac{\theta^2}{4}-\dfrac{π\theta}{2} +\sum_{n=1}^∞ \dfrac{1}{n^2}$ Same process again you will get $\sum_{n=1}^∞ \dfrac{\sin(n\theta)}{n^3} = \dfrac{\theta^3}{12} -\dfrac{π\theta^2}{4} +\theta\sum_{n=1}^∞ \dfrac{1}{n^2}$ $= \dfrac{\theta^3}{12} -\dfrac{π\theta^2}{4} +\dfrac{π^2\theta}{6}$ Put $x=\sqrt{2}$ You will get $\dfrac{\sin(\sqrt{2})}{1^3} +\dfrac{\sin(2\sqrt{2})}{2^3} +\dfrac{\sin(3\sqrt{2})}{3^3} +\cdots= \dfrac{1}{3\sqrt{2}}+\dfrac{π^2}{3\sqrt{2}}-\dfrac{π}{2}$ Answer is $\boxed{2b+a=7}$