AP Calculus BC

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

Q: Why is the limit one rather than zero?

Because \sin h \approx h for small h, so \frac{\sin h}{h} \to 1. The factor of n grows at exactly the rate the sine shrinks, and the two effects cancel.

Q: Can L'Hopital's rule be used here?

On the rewritten form \frac{\sin h}{h}, yes, since it is a genuine 0/0 form. Many courses avoid it, because the standard proof of the derivative of sine already uses this limit.

The series diverges. Writing the term as sine of one over n divided by one over n makes it the classic special trig limit, which equals one. The terms tend to one rather than zero, so the nth term test settles it.

\sum_{n=1}^{\infty} n\sin\!\left(\frac{1}{n}\right)

Diverges

Settled by the nth term test for divergence.

Recognising the special limit

Set $h = 1/n$ , so that $h \to 0$ as $n \to \infty$ . The term becomes $\frac{\sin h}{h}$ , which is the special trig limit every AP student meets in Unit 1.

\lim_{n \to \infty} n\sin\!\left(\frac{1}{n}\right) = \lim_{h \to 0^{+}}\frac{\sin h}{h} = 1 \ne 0

The substitution is what makes this tractable. Left as written, the product looks like infinity times zero, an indeterminate form that says nothing on its own.

Why the indeterminate form matters

A large factor multiplying a shrinking one can tend to 0, to infinity, or to anything in between. Here the two effects balance exactly and the answer is 1, which is precisely the content of the special limit.

Compare $\sum n \sin\!\left(\frac{1}{n^{2}}\right)$ , where the same substitution gives terms behaving like $1/n$ . Those DO tend to 0, so the nth term test goes quiet and limit comparison with the harmonic series is needed instead. It still diverges, but for a different reason.

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 is the limit one rather than zero?

Because $\sin h \approx h$ for small h, so $\frac{\sin h}{h} \to 1$ . The factor of n grows at exactly the rate the sine shrinks, and the two effects cancel.

Can L'Hopital's rule be used here?

On the rewritten form $\frac{\sin h}{h}$ , yes, since it is a genuine $0/0$ form. Many courses avoid it, because the standard proof of the derivative of sine already uses this limit.