Largest $k$ for which $\sum_{n=1}^\infty \frac{\cos nx}{2^n}$ is in $C^k$ with respect to $x$ I was asked to find the largest number $k$ for which the real function $$x \mapsto \sum_{n=1}^\infty \frac{\cos nx}{2^n}$$
is in the differentiability class $C^k$.
I can't seem to think of a way to approach this question..any help will be gladly accepted! 
 A: The series is uniformly convergent on $\mathbb R$ (by the M-test), so the function
$$f(x) = \sum_{n=1}^\infty \frac{\cos nx}{2^n} $$
is differentiable on $\mathbb R$ and
$$f'(x) =- \sum_{n=1}^\infty n\frac{\sin nx}{2^n} $$
You can apply the same trick to $f'$ and conclude that $f'$ is differentiable. By induction you can work out that $f^{(k)}$ is differentiable for all $k$ and so $f \in C^\infty(\mathbb R)$. 
A: While this is probably not the intended method to solve this problem, you can actually sum this series up explicitly via a quick detour into the complex plane and examine the properties of the resulting function.  By Euler's identity, we have
$$
\sum_{n=1}^\infty \frac{\cos nx}{2^n} = \Re \left[ \sum_{n=1}^\infty \frac{e^{inx}}{2^n} \right] = \Re \left[ \sum_{n=1}^\infty \left( \frac{e^{ix}}{2} \right)^n \right].
$$
The sum in square brackets is a geometric series;  and since the magnitude of the ratio of successive terms is
$$
\left| \frac{a_{n+1}}{a_n} \right| = \left|\frac{e^{ix}}{2}\right| = \frac{1}{2} < 1
$$
for all values of $x$, the series converges to
$$
\sum_{n=1}^\infty \left( \frac{e^{ix}}{2} \right)^n = \frac{\frac{1}{2} e^{ix}}{1 - \frac{1}{2} e^{ix}}.
$$
Taking the real part of this, we conclude that
$$
\sum_{n=1}^\infty \frac{\cos nx}{2^n} = \frac{ \frac{1}{2} \cos x - \frac{1}{4}}{\frac{5}{4} - \cos x}
$$
This is obviously a $\mathcal{C}^\infty$ function, since it is the quotient of two $\mathcal{C}^\infty$ functions (and the denominator never vanishes.)
A: A function $f$ is of differentiability  class $C^k$ if the derivatives $f', f'', ...,f^{(k)}$ exist and are continuous, so that is what you would like to check.
By a well known theorem (7.17 in Baby Rudin), if $$f(x) = \lim_{N \to \infty} \underbrace{\sum_{n=1}^N \frac{\cos{nx}}{2^n}}_{f_N(x)}$$ is differentiabile for each $N$, and if the limit exists for some $x$ value, and if $\{f_N'\}$ converges uniformly (for all $x$ values), then
$$f'(x) = \lim_{N \to \infty} F_N'(x).$$
After checking that the "if" conditions are true, you can compute 
$$f'(x) = \lim_{N \to \infty} \frac{d}{dx} \sum_{n=1}^N \frac{\cos{nx}}{2^n} = \lim_{N \to \infty} \sum_{n = 1}^{N} \frac{d}{dx} \frac{\cos{nx}}{2^n} = \lim_{N \to \infty} -\sum_{n = 1}^N \frac{n \sin{nx}}{2^n}.$$ 
Then you can check the "if" conditions again and then compute $f''(x)$, and check again and compute $f'''(x)$, all the way up to $f^{(k)}(x)$ by induction. You can then argue that $f^{(k)}(x)$ is continuous since each term if the sequence is continuous (if $\{f_n\}$ is a sequence of continuous functions and $f_n \to f$ uniformly, then $f$ is continuous by Theorem 7.12 in Baby Rudin).
This is just my initial thoughts, hopefully it gives you something to work with. 
A: Let $$ f: \mathbb{D} \to\mathbb{C} $$ $$f(z) =\sum_{n>0} z ^n $$ and $$f: \mathbb{D} \to\mathbb{C}$$ $$g(z)=\frac{e^{ix}}{2} $$
then $$\sum_{n>0} \frac{\cos nx}{2^n} =\mbox{Re} {(f\circ g)(x)}$$
hence it is harmonic and thus $C^{\infty} $ class.
