Is the sequence $u_{n}=\sum_{p=0}^{n}(\arctan\frac{x}{2^p})^2$ convergent?

Let $x$ be a nonnegative real number. Is the sequence $\displaystyle \left\{\sum_{p=0}^{n}\left(\arctan\frac{x}{2^p}\right)^2\right\}_{n\geq 0}$ convergent ? It's easy to check that it is strictly increasing. If we could prove it is bounded above, the conclusion would follow immediately from Weierstrass theorem (monotone convergent sequence).

-
We have $\lim_{p\to\infty}\frac{\arctan\frac xp}{\frac xp}=1$ hence for $p$ large enough, $\left(\arctan\frac xp\right)^2\leq\frac 32\frac{x^2}{2^{2p}}$. – Davide Giraudo Nov 27 '11 at 9:54

Why not just use a comparison test? $$|\arctan u| \le |u|,\text{ so }(\arctan u)^2 \le u^2.$$
You are right. This allows us to find a real $sup$ for the sequence. – user20010 Nov 27 '11 at 19:30
Put $f_p(x):=\left(\arctan \frac x{2^p}\right)^2$. Then we have $$0\leq f_p(x)=\left(\int_0^{\frac x{2^p}}\frac{dt}{1+t^2}\right)^2\leq\left(\int_0^{\frac x{2^p}}dt\right)^2=\frac{x^2}{2^{2p}},$$ and since $\sum\limits_{p\geq 0}\frac{x^2}{2^{2p}}$ is convergent, the series $\sum\limits_{p\geq 0}f_p(x)$ is convergent for each $x$. The series is normally convergent on each compact but it's not uniformly convergent on $\mathbb R$. In fact, we have $$\sup_{x\in\mathbb R}\sum_{p\geq N}f_p(x)\geq \sum_{p\geq N}f_p\left(2^N\right)\geq \left(\arctan \frac{2^N}{2^N}\right)^2=\frac{\pi^2}{16}$$ for any fixed $N$, hence $\sup_{x\in\mathbb R}\sum\limits_{p\geq N}f_p(x)$ cannot converge to $0$
I had to look at this twice to realize what you're doing in the first two lines. You write about $x/2^p$, but you're actually only proving the inequality $0\le(\arctan u)^2 \le u^2$. I would (1) first say that that's what you're doing, without mentioning $2s/2^p$; then (2) do it; then (3) only after that apply it to $x/2^p$. The way of writing it above gives it a superficial appearance of complexity. – Michael Hardy Nov 27 '11 at 16:10
Typo: I meant $x/2^p$, not $2s/s^p$. – Michael Hardy Nov 27 '11 at 16:17