Is this argument valid or invalid?

Question

For all n>1, $\frac{\arctan(n)}{n^3}$<$\frac{\pi}{2n^3}$, and the series $\frac{\pi}{2}\sum\frac{1}{n^3}$ converges, so by the comparison test, the series $\sum\frac{\arctan(n)}{n^3}$ converges.