Prove $f(x) = \sum_{n=1}^\infty \frac{\sin nx + \cos nx}{n^3}$ is well-defined and $C^1$. 
Prove $f(x) = \sum_{n=1}^\infty \frac{\sin nx + \cos nx}{n^3}$ is well-defined and $C^1$.

First of all I need to prove that $f(x)$ is well-defined. I'm not so sure what does it mean.
Basically I can claim that:
$$ f(x) = \sum_{n=1}^\infty \frac{\sin nx + \cos nx}{n^3} \le \sum_{n=1}^\infty \frac{2}{n^3} < \infty$$
So, by Weierstrass M-test we get that $f(x)$ converges uniformly to some continuous $f(x)$.
Hence, we may take the derivative of the term inside the sum and get:
$$f'(x) = \sum_{n=0}^\infty \frac{n\cos nx -n\sin nx + 3n^2(\sin nx + \cos nx)}{n^6} \le \sum_{n=0}^\infty \frac{2n + 6n^2}{n^6} < \infty$$
So again $f_n'(x)$ continuous and converges uniformly to $f'(x)$ which is also continuous by Weierstrass M-test. 
I'd be glad if you could examine my solution and tell me if it right/rigorous. 

Here's some remarks:


*

*You said that "$f(x)$ converges uniformly to some continuous $f(x)$" which is misspoke and we should write "the series converges uniformly..."

*You said "... Hence, we may take the derivative of the term...": I don't see the implication: how if $f$ is continuous then you may take the derivative?


Finally, notice that to prove $\sum_n f_n(x)$ is well defined and $C^1$ you just need to show:


*

*the series $\sum_n f_n(x)$ is point-wise convergent on $\Bbb R$;

*the functions $f_n$ are $C^1$ on $\Bbb R$;

*the series $\sum_n f'_n(x)$ is uniformly convergent on $\Bbb R$.

