# Show $f(x) = \sum_{n=1}^\infty \frac{\sin{nx}}{n^{5/2}}$ is $C^1$

The task is to show $$f(x) = \sum_{n=1}^\infty \frac{\sin{nx}}{n^{5/2}}$$ is a continuous function on $\mathbb{R}$ with a continuous derivative.

Showing that the series converges at each point is easy, assume that has been shown. Next I suppose I need to show that $f(x)$ is continuous. My thought was to define $$f_N(x) = \sum_{n=1}^N \frac{\sin{nx}}{n^{5/2}},$$ and show that $f_N \to f(x)$ uniformly (then $f$ being continuous follows).

I think the following is a valid proof that $f_N \to f$ uniformly. It suffices to show $|f_N(x) - f_M(x)| < \epsilon$ for all $x$ when $N,M >$ some $Z \in \mathbb{Z},$ and assume $N > M.$ $$|f_N(x) - f_M(x)| = \sum_{n=M+1}^N \frac{\sin{nx}}{n^{5/2}} < \sum_{n=M+1}^N n^{-2} < \epsilon$$ for large enough $M,N.$

The part I am stuck on is showing $f'(x)$ exists for all $x \in \mathbb{R}.$ Rudin's book states that if

$f_n$ are differentiable
$f'_n$ uniformly converge
$f_n(x_0)$ converges for some point $x_0$
Then $f_n \to f$ uniformly and $f' = \lim f'_n(x)$

I am wondering if there is an easier way to show that $f(x)$ is continuously differentiable? And if not, would showing $f'_n$ converge uniformly be the same as above - using a Cauchy sequence for $f'_n$'s?

• As another user190080 noted, you may use the Weierstrass M-test to prove uniform convergence. You have essentially provided a proof for this fact via Cauchy's criterion for uniform convergence with $M_n=n^{-2}$. Jul 14, 2016 at 16:40

Actually you could use the Weierstrass-M-criterion to show that the above sequence is converging uniformly, just observe that $$\bigg|\frac{\sin{(nx)}}{n^{5/2}}\bigg|\leq\frac{1}{n^{5/2}}$$ and furthermore $$\sum_{n=1}^{\infty}\frac{1}{n^{5/2}}=\xi(\frac 5 2)<\infty$$ hence while using the above criterion you can infer that the sequence $$f_N(x) = \sum_{n=1}^N \frac{\sin{nx}}{n^{5/2}}$$ converges absolutely and uniformly to some $f$.