Evaluate $\sum_{n=1}^\infty\frac{\sin(n)}{n^3}$

The best way to compute such series is using periodic functions. Suppose f is periodic with period $2\pi$ and is equal to ${x^{3}\over3}-{{\pi x^{2}\over 2}}+{{\pi ^{2}x}\over 6}$ in the interval $[-\pi,\pi]$. The Fourier series of this function is $\Sigma_{n=1}^{\infty} {\sin(2nx)\over n^{3}}$. Therefore replacing x=0.5 the result jumps out!

You can not apply the approach from that paper directly as the key assumption of theorem 3.2 $$ |f(z)|\le \frac{M}{|z|^k} $$ for some $k>1$ and all large $z$, $|z|>R$ for some large $R$, is not satisfied for $f(z)=\frac{\sin z}{z^3}$ as the sine function is exponential in the imaginary directions, $$ \sin(z)=\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y) $$