What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes to infinity. 
I originally thought the resulting series can be written this way:
$$(1 + \frac{1}{2} + \frac{1}{3} ...) - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^3}{3} + \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^5}{5} - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^7}{7} ...$$
But that is wrong. The right way to "insert" the series is:
$$1 + \frac{1}{2} + \frac{1}{3} ... - \frac{1}{3} - \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{3})^3}{3} ... + \frac{1}{5} + \frac{(\frac{1}{2})^5}{5} + \frac{(\frac{1}{3})^5}{5} ... ...$$
So it looks like a bunch of harmonic serieses manipulated. Harmonic series diverges, but I remember that doesn't necessarily mean a similar series diverges. I remember from Calculus 2 that, for example, $\lim_{n\to\infty}$ of $\sin(n)\over n$ converges to zero even though $\sin(n)$ does not converge.
So how do I figure out what this series converges to, if anything?

It seems it has not yet been pointed out that your rearrangement does not work!

$\frac11 - \frac12 + \frac13 - \frac14 + \cdots = \ln(2) > \frac12$.
$\frac11 - \frac12 + \frac13 - \frac14 + \cdots \ne \color{red}{( \frac11 + \frac13 + \frac15 + \frac17 + \cdots ) - ( \frac12 + \frac14 + \frac16 + \frac18 \cdots )}$.   [RHS is ill defined!]
$\frac11 - \frac12 + \frac13 - \frac14 + \cdots $
$\ \ne ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac13 ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots )$
$\ \quad + \frac1{3^2} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac1{3^3} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \cdots$
$\ = \frac12 \times ( \frac1{1 \times 2} - \frac1{4 \times 5} + \frac1{7 \times 8} - \cdots ) < \frac14$.

