$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^3+x}$
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.
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Sign up to join this community$\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^3+x}$
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.
Hint: it is sufficient to show that the derivatives converge uniformly and that original series converges at (at least) one point. To show uniform convergence, consider using the Weierstrass M-test.
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 $\sum M_n < \infty$.