AP Calculus BC

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

The sum of sin n over n squared converges, and converges absolutely. The absolute value of sin n never exceeds 1, so the absolute terms are at most 1 over n squared, and that p-series converges. Direct comparison finishes it.

∑n=1∞sin⁡nn2\sum_{n=1}^{\infty}\frac{\sin n}{n^{2}}

Converges

Settled by the direct comparison test.

Bound the numerator, then compare

The only fact needed about the numerator is that sine is bounded, which holds no matter how erratic sin⁡n\sin n looks at integer inputs.

∣sin⁡nn2∣≤1n2\left|\frac{\sin n}{n^{2}}\right| \le \frac{1}{n^{2}}

Since ∑1n2\sum \frac{1}{n^{2}} converges as a pp-series with p=2p = 2, direct comparison gives convergence of the absolute series, so the original converges absolutely.

Comparison needs positive terms

The direct comparison test requires nonnegative terms, and sin n changes sign constantly. Taking absolute values first is what makes the test legal, and it is why the conclusion comes out as ABSOLUTE convergence.

Why the signs do not matter here

The terms are not alternating in any regular pattern, because sin⁡n\sin n at integer nn jumps around unpredictably. That rules out the alternating series test entirely.

Absolute convergence is the tool that does not care. Once the absolute series converges, the original converges no matter how the signs fall.

The mistakes students make

  • Applying the alternating series test. The signs do not alternate regularly, so it does not apply.
  • Trying to evaluate lim⁡sin⁡n\lim \sin n. It does not exist, and the argument never needs it.
  • Comparing without absolute values. Direct comparison needs nonnegative terms.

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 sin(n)/n^2 converge?

Yes, absolutely, by comparison with ∑1n2\sum \frac{1}{n^{2}}.

Does sin(n) have a limit?

No. At integer inputs it never settles, which is precisely why the argument bounds it instead of evaluating it.