Proofs of differentiability

Question

How can I prove that

$$\ f(x)= \sum_{n=1}^\infty \frac{x^n}{2^n}\cos(nx) $$

defines function differentiable in (-2, 2)? I have to evaluate the derivative at $\ 0$.

My idea is to check the radius of convergence, within which it is uniformly convergent. Then I check the sum of derivatives whether it is convergent - then I see whether it is convergent. However I have no idea how to deal with cosine. I would appreciate if you would explain to me how to solve this.

Answer 1

If $x \in (-2,2)$ then $\left| \dfrac{x^n}{2^n} \cos(nx) \right| \leq \dfrac{|x|^n}{2^n} \cdot 1 = \left(\dfrac{|x|}{2} \right)^n < 1$. In other words, the series you have is dominated in absolute value by a convergent geometric series. So, the series converges absolutely for all $x \in (-2,2)$.

Answer 2

Hint: Let $F_N$ be the $N$'th partial sum. $F_N'(x)$ converges to a continuous function on $(-2,2)$, convergence being uniform on $[-2+\epsilon, 2-\epsilon]$ for any $\epsilon$. Note that $F_N(x+h) - F_N(x) = \int_x^{x+h} F_N'(t)\; dt$.