Is this sum equivalent?

Question

Does $\sum_{n=1}^\infty -\frac{4\cos(nx)}{10n^3 \pi}\cos(nt)\sin(nx)=\frac{2}{n^3}\cos(nt)\sin(nx)(1-\cos(nx))$?

I don't think it does. How can I check?

Answer 1

One possible form for the series is \begin{align} \sum_{n=1}^{\infty} \frac{\sin(2n x) \cos(n t)}{n^{3}} = \frac{5 \pi^{2}}{16} \, (t+2x)^{2} \left[ H(2x) - H(-2x) \right] - \frac{5 \pi^{2}}{16} \left[ (t+2x)^{2} H(t+2x) - (t-2x)^{2} H(t-2x) \right] \end{align} where $H(x)$ is the Heaviside step function.

Answer 2

If $\sum_{n=1}^\infty -\frac{4\cos(nx)}{10n^3 \pi}\cos(nt)\sin(nx)$ exists, then it is a function $f(x,t)$, however, $\frac{2}{n^3}\cos(nt)\sin(nx)(1-\cos(nx))$ is a function $g(n,x,t)$.

It is imposibble that they are equivalent.