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?

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 $0.4007$.
