How do I determine whether or not $ \sum_{n=1}^{\infty} \left( 1-\frac{2}{\pi}\arctan{n} \right) $ converges? $$
\sum_{n=1}^{\infty} \left( 1-\frac{2}{\pi}\arctan{n} \right)
$$
Since $\arctan{x}$ is increasing and $\lim_{x\to \infty} \arctan{x}=\frac{\pi}{2}$, all the terms of the series are nonnegative. Also, it satisfies the necessary condition of convergence. Can we use theorems in Baby Rudin to solve this problem? I would appreciate if you could give me some hints.
 A: Using @Mourad comment, that is to say
$$\arctan\left(n\right)=\frac{\pi}{2}-\arctan\left(\frac{1}{n}\right)$$ your problem is to compute first the partial sum
$$S_p=\frac2{\pi} \sum_{n=1}^p\arctan\left(\frac{1}{n}\right)=1+\frac2{\pi} \sum_{n=4}^p\arctan\left(\frac{1}{n}\right) $$ Now, use the series expansion,
$$\arctan\left(\frac{1}{n}\right)=\frac{1}{n}-\frac{1}{3 n^3}+\frac{1}{5 n^5}+O\left(\frac{1}{n^7}\right)$$ which, for $n=4$ would give $\frac{3763}{15360}\approx 0.244987$ while $\arctan\left(\frac{1}{4}\right)\approx 0.244979$.
This would give, as an approximation,
$$S_p=1+\frac2{\pi}\left(H_p-\frac{H_p^{(3)}}{3}+\frac{H_p^{(5)}}{5}-\frac{64271}{38880} \right)$$ Using the asymptotics of generalized harmonic numbers, this would give
$$S_p=1+\frac2{\pi}\left(\log (p)+\frac{\zeta (5)}{5}-\frac{\zeta (3)}{3}+\gamma -\frac{64271}{38880}\right)+O\left(\frac{1}{p}\right)$$
$$S_p \approx 0.192037+\frac2{\pi} \log(p)$$ For $p=100$, the approximation would give $3.12378$ for an exact value equal to $3.12696$.
A: The series is divergent. By the comment of Mourad we can write this as $\frac  2 {\pi}\sum \arctan (\frac  1 n)$. By L'Hopital's Rule $\frac {\arctan x } x \to 1$ as $x \to 0$. Hence $\arctan x >\frac   x  2$ for positive $x$ sufficiently small. Now compare with the series $\sum \frac  1 {2n}$.
