# converge of the series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^3+x}$ to continuously differentiable function

$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^3+x}$

show that the series converge in $(-1,\infty)$ to a continuously differentiable function.

I know it's something with uniformly convergence but not sure how.

• So you didn't try anything? Weierstrass M? The theorem on uniform convergence and differentiation? Hard to believe you have no idea here ...
– zhw.
Jul 17, 2018 at 19:46

Here's a further hint for uniform convergence: the derivative of the $n$th term is $$-\frac{n\sin(nx)}{n^3 + x} + \frac{\cos(nx)}{(n^3 + x)^2}\,.$$
In order to show uniform convergence, it is sufficient to bound this term above for $x \in (-1,\infty)$ by a constant $M_n$ so that $\sum M_n < \infty$.