This is Exercise 15 Chapter 3 from Stein's Fourier Analysis.
$$f(x)=\sum_{k=0}^\infty 2^{-k\alpha}e^{i2^k x},$$ where $0<\alpha<1$, satisfies
$$|f(x+h)-f(x)|\le C|h|^{\alpha}.$$
[Hint: break up the sum as follows $f(x+h)-f(x)=\sum_{2^k \le 1/|h|}+\sum_{2^k >1/|h|}$. To estimate the first sum use the fact that $|1-e^{i\theta}|\le |\theta|$ whenever $\theta$ is small. To estimate the second sum, use the obvious inequality $|e^{ix}-e^{iy}|\le 2.$]
I used the hint to obtain $$|f(x+h)-f(x)|\le \sum_{k=0}^\infty 2^{-k\alpha}|e^{i2^k h}-1| \le \sum_{2^k \le 1/|h|} 2^{-k\alpha}2^k |h| + \sum_{2^k > 1/|h|} 2^{-k\alpha} 2.$$
I have trouble with the second sum. I can see that the terms in the last sum are bounded by $2|h|^{\alpha}$, however, since the sum contains infinitely many $k$'s how can I conclude from this that the sums are bounded by a constant multiple $|h|^{\alpha}$?
Let $k_0$ be the smallest positive integer $k$ such that $2^k > 1/\lvert h\rvert$. Then the sum of the series $\sum\limits_{2^k > 1/\lvert h\rvert} 2^{-k\alpha}$ is less than the sum of $\sum\limits_{k = k_0}^\infty 2^{-k\alpha}$, which is $c_\alpha2^{-k_0 \alpha}$, where $c_\alpha = \dfrac{1}{1-2^{-\alpha}}$. Since $2^{k_0} > 1/\lvert h\rvert$, then $2^{-k_0\alpha} < \lvert h\rvert^{\alpha}$. So the sum of the series is less than $c_\alpha\lvert h\rvert^{\alpha}$.