AP Calculus BC

Does the Sum of arctan(n)/n^2 Converge? Yes

Q: Does the sum of arctan(n)/n^2 converge?

Yes, by comparison with \frac{\pi}{2}\sum \frac{1}{n^{2}}.

The sum of arctan n over n squared converges. Arctangent never exceeds pi over 2, so each term is at most pi over 2 times 1 over n squared, and a constant multiple of a convergent p-series still converges.

\sum_{n=1}^{\infty}\frac{\arctan n}{n^{2}}

Converges

Settled by the direct comparison test.

Bound the numerator

Arctangent increases toward a horizontal asymptote, so it is bounded for every $n$ .

0 < \arctan n < \frac{\pi}{2} \implies \frac{\arctan n}{n^{2}} < \frac{\pi}{2}\cdot\frac{1}{n^{2}}

The right-hand side is a constant multiple of a convergent $p$ -series, so direct comparison gives convergence.

Any bounded numerator behaves the same way

Replace arctan with sine, cosine, or anything that stays inside fixed limits and the argument is unchanged. It is the boundedness doing the work, not the particular function.

The mistakes students make

Trying to evaluate $\lim \arctan n$ and getting stuck. The argument needs only a bound, and the limit happens to be $\frac{\pi}{2}$ .
Forgetting the constant. Multiplying a convergent series by $\frac{\pi}{2}$ leaves it convergent.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of arctan(n)/n^2 converge?

Yes, by comparison with $\frac{\pi}{2}\sum \frac{1}{n^{2}}$ .

Does the bound have to be tight?

No. Any finite bound on the numerator works, because a constant multiple never changes convergence.