Prove $\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}$ is continuous and uniformly continuous on $\mathbb{R}$ Let $$f(x) = \sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}$$ 
a) Prove $f$ is continuous on $(-\infty, \infty)$
b) Is $f$ uniformly continuous? Justify your answer. 
I know that $f$ is the sum of continuous functions of $x$, and so it is continuous. How can I more rigorously prove this? Here's what I have. 
$|\sum_{n=1}^{\infty} \frac{sin(nx)}{1+n^3}| \leq |\sum_{n=1}^{\infty} \frac{1}{1+n^3}| < |\sum_{n=1}^{\infty} \frac{1}{n^3}| =|\zeta(3)|$
Any help would be appreciated. 
 A: The given series converges uniformly on $(-\infty,\infty)$ to $f$, since each term is a continuous function on $(-\infty,\infty)$ thus $f$ is a continuous function on $(-\infty,\infty)$. The uniform convergence of the series may be obtained from the normal convergence:
$$
\left|\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}\right| \leq \sum_{n=1}^{\infty} \frac{\left|\sin(nx)\right|}{1+n^3} \leq \sum_{n=1}^{\infty} \frac1{1+n^3}<\infty. 
$$ On the other hand, from
$$
\left|\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}-\sum_{n=1}^{\infty} \frac{\sin(ny)}{1+n^3}\right| \leq \sum_{n=1}^{\infty} \frac{\left|2 \cos \left( {{nx+ny} \over 2} \right) \sin \left( {{nx-ny} \over 2} \right)\right|}{1+n^3} \leq \underbrace{\left(\sum_{n=1}^{\infty} \frac{n}{1+n^3}\right)}_{<\infty}|x-y|, 
$$ where we have used $|\sin x|\leq |x|$, we deduce that  $f$ is uniformly continuous on $(-\infty,\infty)$.
A: a)  Note: $|f_n(x)| = \Bigg| \frac{\sin(nx)}{1+n^3}\Bigg| \leq \Bigg|\frac{1} {1+n^3} \Bigg|$ for all $n \in \mathbb{N}$ and for all $x \in \mathbb{R}$. Observe also that $\sum_{n=1}^{\infty} \frac{1} {1+n^3}\leq \infty$. 
So by the Weierstrass M-Test, $\sum_{n=1}^{\infty} \frac{\sin(nx)}{1+n^3}$ converges uniformly on $\mathbb{R}$. 
Note also that $f_n = \frac{\sin(nx)}{1+n^3}$ is uniformly continuous for all $n \in \mathbb{N}$. Then it follows from the Term -by-term Continuity Theorem (Abbott 2nd edition page 188) that $f(x)$ is continuous on $\mathbb{R}$. 
b) A uniform limit on a set $I$ of uniformly continuous functions on $I$
is uniformly continuous on $I$.
