TITLE: How to justify an infinite series being continuous and having continuous derivative
QUESTION [0 upvotes]: Question
Consider the series $g(x)=\sum_{i=1}^{\infty}$$2^{-k}cos(kx)$ and show that it defines a continuous series from R$\rightarrow$R and has a continuous derivative.
Attempt
I know that each of the individual terms are continuous, and that, in general, the sum of continuous functions on the same interval should be continuous. However, since this is an infinite sum, I am hesitant to make this argument. It also feels difficult to apply this to epsilon-delta definition of continuity to make any meaningful conclusion about the sum.
By this same reasoning, I am unsure about the derivative. I know that term by term differentiation is justified by the M-test, however, I'm sure about how to go about formulating a continuity argument here.

REPLY [3 votes]: If $k\in\Bbb N$, let $g_k(x)=2^{-k}\cos(kx)$. Then $g_k'(x)=-k2^{-k}\sin(kx)$, and therefore $|g_k'(x)|\leqslant k2^{-k}$. Since the series $\sum_{k=1}^\infty k2^{-k}$ converges, it follows from the Weierstrass $M$-test that the series $\sum_{k=1}^\infty g_k'$ converges uniformly and, since each $g_k'$ is a continuous function, its sum is a continuous function. But then $\sum_{k=1}^\infty g_k$ is differentiable and$$\left(\sum_{k=1}^\infty g_k\right)'=\sum_{k=1}^\infty g_k',\tag1$$which is continuous. And, since $\sum_{k=1}^\infty g_k$ is differentiable, it is a continuous function.
The theorem that I have used to deduce that $\sum_{k=1}^\infty g_k$ is differentiable and that $(1)$ holds was the theorem that states that if a sequence $(f_n)_{n\in\Bbb N}$ of differentiable functions converges pointwise to a function $f$ and if $(f_n')_{n\in\Bbb N}$ converges uniformly to a function $\varphi$, then $f$ is differentiable, and $f'=\varphi$.
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