AP Calculus BC

Does the Sum of cos(1/n) Converge? No

The series diverges. As n grows the argument one over n shrinks toward zero, and the cosine of something near zero is near one. The terms therefore tend to one rather than zero, and the nth term test settles it.

∑n=1∞cos⁡ ⁣(1n)\sum_{n=1}^{\infty}\cos\!\left(\frac{1}{n}\right)

Diverges

Settled by the nth term test for divergence.

Follow the argument, not the function

The instinct is to see a cosine and expect oscillation. The oscillation is in the ARGUMENT's behaviour, and here the argument 1/n1/n is marching quietly to 0, not sweeping across the circle.

lim⁡n→∞cos⁡ ⁣(1n)=cos⁡0=1≠0\lim_{n \to \infty}\cos\!\left(\frac{1}{n}\right) = \cos 0 = 1 \ne 0

Cosine is continuous, so the limit passes inside: the limit of the cosine is the cosine of the limit. Every term past n=5n = 5 already exceeds 0.980.98.

Contrast with cos n

The series ∑cos⁡n\sum \cos n is a genuinely different animal. There the argument grows without bound and the terms swing between −1-1 and 1 forever, so the limit of the terms does not exist. That also gives divergence by the nth term test, but by failure to exist rather than by tending to a nonzero value.

Both fail the same hypothesis, which is why the nth term test is stated as terms failing to tend to 0 rather than terms tending to something nonzero.

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 nth term test apply when the limit does not exist?

Yes. The hypothesis is that the terms fail to tend to 0, and never settling is one way to fail. ∑(−1)n\sum (-1)^{n} and ∑cos⁡n\sum \cos n both diverge for that reason.

What about the sum of sine of one over n?

That one is different. sin⁡(1/n)→0\sin(1/n) \to 0, so the nth term test is silent, and limit comparison with ∑1/n\sum 1/n gives divergence instead.