AP Calculus BC

Does the Sum of 1/(n^2-1) Converge? Yes

The sum of 1 over n squared minus 1, starting at n equals 2, converges. Direct comparison with 1 over n squared fails because these terms are LARGER, so the limit comparison test is used instead: the ratio tends to 1, so the two series share a verdict.

∑n=2∞1n2−1\sum_{n=2}^{\infty}\frac{1}{n^{2}-1}

Converges

Settled by the limit comparison test.

Why direct comparison fails here

Subtracting from the denominator makes each term BIGGER, so the inequality points the wrong way for proving convergence.

1n2−1>1n2\frac{1}{n^{2}-1} > \frac{1}{n^{2}}

Being larger than a convergent series proves nothing. Limit comparison does not need an inequality at all, only a finite positive ratio.

lim⁡n→∞1n2−11n2=lim⁡n→∞n2n2−1=1\lim_{n \to \infty}\frac{\frac{1}{n^{2}-1}}{\frac{1}{n^{2}}} = \lim_{n \to \infty}\frac{n^{2}}{n^{2}-1} = 1

A limit of 11 is finite and positive, so both series do the same thing, and ∑1n2\sum \frac{1}{n^{2}} converges.

It also telescopes

Partial fractions gives an exact sum, which is unusual for a comparison problem.

1n2−1=12(1n−1−1n+1)\frac{1}{n^{2}-1} = \frac{1}{2}\left(\frac{1}{n-1} - \frac{1}{n+1}\right)

The partial sums collapse to 12(1+12−1N−1N+1)\frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N} - \frac{1}{N+1}\right), so the total is 34\frac{3}{4}.

The mistakes students make

  • Using direct comparison with 1n2\frac{1}{n^{2}} anyway. The inequality runs the wrong way and proves nothing.
  • Starting at n=1n = 1, where the denominator is 00.

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 1/(n^2-1) converge?

Yes, by limit comparison with ∑1n2\sum \frac{1}{n^{2}}, whose ratio is 11.

Why not direct comparison?

Because these terms are LARGER than 1n2\frac{1}{n^{2}}, and being larger than a convergent series proves nothing.

Why start at n = 2?

At n=1n = 1 the denominator is zero.