What does this series converge 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.

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

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 $0.4007$.