Con(/di)vergence of $\sum \tan^{-1}(\frac 1n)$ via comparison test My friend showed me a problem yesterday that I couldn't figure out.  He's supposed to decide whether $$\sum_{n=2}^\infty \tan^{-1} \left(\frac 1n\right)$$ converges or divergences by using the comparison test.  My first thought was to try to compare it to one of its Taylor polynomials, but he tells me he hasn't gotten to Taylor expansions yet.  So I can't come up with anything to compare this to.
 A: You can solve this problem by doing a limit comparison with $\frac{1}{n}$ since we know that $\sum_{n=1}^\infty{\frac{1}{n}}$ diverges by the P-test.
Thus $\sum_{n=2}^\infty{\tan^{-1}(\frac{1}{n})}$ will diverge if
$$
\lim_{n \to \infty}{\tan^{-1}(\frac{1}{n}) \over \frac{1}{n}} \not = 0 \; \text{or}\; \pm \infty.
$$
Notice that this satisfies L'Hopitals, the numerator and denominator both go to 0. So we can take the derivative of the top and the bottom to see:
$$
\lim_{n \to \infty}{\tan^{-1}(\frac{1}{n}) \over \frac{1}{n}} = \lim_{n \to \infty}{\left(\frac{1}{1 + \left(\frac{1}{n}\right)^2} \right) \cdot \frac{-1}{n^2} \over \frac{-1}{n^2}} = \lim_{n \to \infty}{\frac{1}{1 + \left(\frac{1}{n}\right)^2}} = 1.
$$
Because the limit equals one, we know by comparison test that $\sum_{n=2}^\infty{\tan^{-1}(\frac{1}{n})}$ diverges.
Edit: For problems similar to this, the best place to start is by trying to compare it with the function inside the trig function, the main reason for that is that when using L'Hopitals (if appropriate) will have the derivative of the inside (by chain rule) cancel with the derivative in the denominator as shown above.
A: Claim: $\tan^{-1}\left(\frac{1}{n}\right)\geq\frac{1}{2n}$ for $n$ sufficiently large.
This is equivalent to $\frac{1}{n}\geq\tan\left(\frac{1}{2n}\right)$.  Observe that $\tan\left(\frac{1}{2n}\right)=\frac{\sin\left(\frac{1}{2n}\right)}{\cos\left(\frac{1}{2n}\right)}\leq \frac{1}{2n}\cdot\frac{1}{\cos\left(\frac{1}{2n}\right)}$.  For $n$ sufficiently large, $2\cos\left(\frac{1}{2n}\right)>1$, so $\tan\left(\frac{1}{2n}\right)$ is at most $\frac{1}{n}$.
Then, you can use a direct comparison with half the harmonic series.
A: By MVT
$$\arctan(\frac 1n)-\arctan(0)=\frac 1n \frac{1}{1+c^2}>\frac 1n\frac{1}{1+1}$$
since $0<c<\frac 1n \leq 1$
$\implies$ by comparison test, 
$\sum\arctan(\frac 1n )$ diverges.
