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This is related to a previous question where, as stated there, $f_{2}(n)$ gives the greatest power of $2$ that divides $n$. Specifically the sequence $\lbrace 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,\cdots\rbrace$ is given by the formula below at the integers $$\displaystyle\begin{align} f_2(n)=\sum_{r=1}^{\infty}\frac{r}{{2^{r+1}}}\sum_{k=0}^{2^{r+1}-1}\cos\left( \frac{2k\pi(n+2^{r})}{2^{r+1}} \right)\end{align}$$

The formula above can be reduced to the formula below. I've been looking at $f_{2}(x)$ and wondering if it converges everywhere

$$ \begin{align*} f_{2}(x)=\lim_{m \to +\infty}f_{2,m}(x)= \lim_{m \to +\infty}\sum_{r=1}^{m}r \cdot \frac{\text{sinc}(\pi(x+2^{r}))}{\text{sinc}(\frac{\pi(x+2^{r})}{2^{r+1}})}\cdot \cos\left(\frac{\pi(x+2^{r})(2^{r+1}-1)}{2^{r+1}} \right ) \end{align*} $$ I made some calculations for $m=16,32,64,128$ and the graphic below is for $m=256$:

Calculations for $m=256$

And some comparative plotting for $m=4$ and $m=16384$, from $x=-1$ to $x=23$:

plotting for m=4 and m=16384

plotting for m=16384 and m=65536

By visual inspection I can't see any difference as I increase the value of $m$. So I would say that $\lim_{m \to +\infty}f_{2,m}(x)$ converges for $x \in \mathbb{R}$, but I'm looking for a trustable proof. I'm also interested is the case when $x \in \mathbb{C}$.

How to prove (or disprove) the convergence of $f_{2}(x)$?


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What is the sequence you are saying that converges? What convergence do you want pintwisely, uniform ? – checkmath May 2 '12 at 12:09
Yes, uniform. The sequence is $\{0,1,0,2,0,1,0,3, \dots\}$ for $f_{2}(n)$ with $n \in \mathbb{N}$ and does not converge I mean the convergence of $\lim_{m \to +\infty}f_{2,m}(x)$ for values of $x \in \mathbb{C}$. – Neves May 2 '12 at 13:38

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