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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).

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    We have $\lim_{p\to\infty}\frac{\arctan\frac xp}{\frac xp}=1$ hence for $p$ large enough, $\left(\arctan\frac xp\right)^2\leq\frac 32\frac{x^2}{2^{2p}}$.2011-11-27

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Why not just use a comparison test? $ |\arctan u| \le |u|,\text{ so }(\arctan u)^2 \le u^2. $

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    You are right. This allows us to find a real $sup$ for the sequence.2011-11-27
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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 $0$

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    @MichaelHardy: you' re right, in fact I should begin like in your answer. If I modify mine, these ones will be too similar.2011-11-27