Is $F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos (nt)}{2^n} \text{d} t$ continuous in $\mathbb{R}$? Is my reasoning for whether $$F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos (nt)}{2^n} \text{d} t$$ is continuous in $\mathbb{R}$ correct?:
Proof
I claim it is continuous in $\mathbb{R}$. $\sum_{0}^{\infty}\frac{\cos(nx)}{2^n}$ is uniformly convergent in $\mathbb{R}$ (and the functions in the series are continuous), therefore  $$F(x)=\int_{0}^{x}\sum_{0}^{\infty}\frac{\cos(nt)}{2^n} \text{d} t = \sum_{0}^{\infty} \int_{0}^{x}\frac{\cos(nt)}{2^n} \text{d} t=x+\sum_{1}^{\infty} \frac{\sin(nt)}{2^n\cdot n}\; .$$ 
$\sum_{1}^{\infty} \frac{\sin(nt)}{2^n\cdot n}$ is uniformly convergent in $\mathbb{R}$, so $F(x)$ is a uniformly convergent series of continuous functions, which means it is continuous.
Have I correctly solved this exercise? Thank you!
 A: It looks good, but some additional justifications should be made. At the start (and for the later series) you should state why the series converges uniformly (by the Weierstrass_M-test, e.g.).  
Also, to justify that the sum of the series is integrable and that switching the order of summation and integration is valid, you should state that the terms of $G(x)=\sum\limits_{n=0}^\infty {\cos(nt)\over 2^n}$ are integrable over any interval $[0,x]$. 
But, can save a few steps in your argument. Since $\sum\limits_{n=0}^\infty {\cos(nt)\over 2^n}$ converges uniformly to $G(x)$, and since the terms of this series are continuous, $G(x)$ is a continuous function. The 
Fundamental Theorem of Calculus immediately gives you the continuity of $F(x)=\int_0^x G(t)\,dt$.
A: You could use the M-test as follows.
$$ \displaystyle \sum\limits_{n = 0}^\infty  {\left|\frac{{\cos nt}}{{{2^n}}}\right|}  < \sum\limits_{n = 0}^\infty  {\frac{1}{{{2^n}}}}  < 2$$
and do the same for the $ \sin tn $ series to make your proof complete.
Otherwise, the proof is correct.
