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?
 A: 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 $
$\ \ne ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac13 ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots )$
$\ \quad + \frac1{3^2} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \frac1{3^3} ( \frac11 - \frac12 - \frac14 + \frac15 + \frac17 - \frac18 - \cdots ) + \cdots$
$\ = \frac12 \times ( \frac1{1 \times 2} - \frac1{4 \times 5} + \frac1{7 \times 8} - \cdots ) < \frac14$.

A: We have,
$$\lim_{n \to \infty} \frac{\arctan(\frac{1}{n})}{\frac{1}{n}}=1$$
This follows easily by first the change of variables $\frac{1}{n}=h$ then by Taylor series.
We also have that $a_n=\arctan (\frac{1}{n}) \geq 0$ for all $n \geq 1$. Hence by the limit comparison test  your series $\sum_n a_n$ diverges with comparison to the harmonic series.
A: Based on this (On the arctangent inequality.) answer, we have:
$$\frac{\arctan x}{x} \geq 1/2$$
for $x \in (0,1]$.
So letting $x = \frac{1}{n}$, we have $\arctan \frac{1}{n} \geq \frac{1}{2n}$ for each $n\geq 1$, so by the comparison test, your series diverges
A: For $0 < x \leq 1$, the series
$$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ... $$
is an alternating series with decreasing terms. This means the terms alternate between overshooting and undershooting the actual value of $\arctan(x)$. In particular, 
$$ 0 < x \leq 1 \implies x - \frac{x^3}{3} < \arctan(x) < x $$
We can use this to get a good bound on the partial sums
$$ \sum_{k=1}^n \arctan\left( \frac{1}{k} \right) 
<  \sum_{k=1}^n \frac{1}{k} = H_n$$
$$  \sum_{k=1}^n \arctan\left( \frac{1}{k} \right) 
> \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{3 k^3} \right) 
= H_n -
\sum_{k=1}^n \left( \frac{1}{3 k^3} \right)  
> H_n - \frac{1}{3} \zeta(3)$$
So not only does the infinite sum go to infinity, it does so in basically the same fashion as harmonic numbers $H_n$ do, and furthermore the error in this estimate is strictly less than $0.4007$.
