The asymptotic behavior of $\sum_{n=1}^\infty\frac{1-\cos(x4^n)}{2^n}$ as $x\to 0$

Question

Is there a way to show that for small $x$'s

$$\sum_{n=1}^\infty\frac{1-\cos(x4^n)}{2^n}\le c\sqrt x$$

I tried Taylor expansion of $\cos$ and square root...

Thank's

Answer 1

Taylor expansion of $\cos x$ is very relevant. Taylor expansion of square root, not at all (it's not an expansion at $0$, is it?)

For small $n$, the numerator is small because $1-\cos t=O(t^2)$; after that, the best we can do is bound it by $2$. The transition happens when $4^n x$ reaches, say, $1$ (precise number does not matter). That is, around $n = \log_4(1/x)$.

Small $n$

$$ \sum_{n\le \log_4 (1/x)} \frac{1-\cos 4^n x}{2^n} \le \sum_{n\le \log_4 (1/x)} C\frac{(4^n x)^2}{2^n} = Cx^2 \sum_{n\le \log_4 (1/x)} 4^{3n/2} $$ and since the geometric sum is comparable to its largest term, this is comparable to $$ x^2 4^{\frac32\log_4 (1/x)} =x^{1/2}$$

Large $n$

$$ \sum_{n\ge \log_4 (1/x)} \frac{1-\cos 4^n x}{2^n} \le \sum_{n\ge \log_4 (1/x)} \frac{2}{2^n} \sim 2^{-\log_4(1/x)} \sim x^{1/2} $$

You can put absolute values around everything if you want.