AP Calculus BC

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

Q: Does the same argument work for cosine n over n?

No. The bound gives \sum 1/n, which diverges, so the comparison fails. That series does converge, but the proof needs tools beyond the AP course.

The series converges, and absolutely. The cosine never exceeds one in absolute value, so the size of each term is at most one over n squared, and that p-series converges. Bounding an oscillating numerator is the whole method.

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

Converges

Settled by the direct comparison test.

Bound the numerator, then compare

The terms are not positive, and they do not alternate in any regular pattern either, since $\cos n$ wanders as $n$ runs through the integers. Neither the comparison test nor the alternating series test applies to the series as written.

\left|\frac{\cos n}{n^{2}}\right| \le \frac{1}{n^{2}}

So compare the ABSOLUTE series instead, which is positive and dominated by a convergent p-series. Absolute convergence then hands back ordinary convergence.

Why this pattern is worth memorising

Any bounded numerator over a convergent p-series denominator converges absolutely: $\frac{\sin n}{n^{2}}$ , $\frac{\arctan n}{n^{2}}$ and $\frac{(-1)^{n}}{n^{2}}$ all work the same way. The numerator's behaviour is irrelevant once it is bounded.

The denominator is what does the work. Change it to $n$ and the argument collapses: $\sum \frac{\cos n}{n}$ actually does converge, but proving it needs Dirichlet's test rather than a bound, and that is well beyond AP.

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

Why take absolute values first?

Because the comparison test requires non-negative terms. Bounding the sizes gives a positive series to compare, and absolute convergence implies convergence for the original.

Does the same argument work for cosine n over n?

No. The bound gives $\sum 1/n$ , which diverges, so the comparison fails. That series does converge, but the proof needs tools beyond the AP course.