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?

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.
