AP Calculus BC

Does the Sum of 1/(3n+2) Converge? No

The series diverges. Limit comparison with the harmonic series gives one third, a finite nonzero number, so the two share a fate. Scaling the harmonic series by any positive constant leaves it divergent.

n=113n+2\sum_{n=1}^{\infty}\frac{1}{3n+2}

Diverges

Settled by the limit comparison test.

The limit comparison

limn1/(3n+2)1/n=limnn3n+2=13\lim_{n \to \infty}\frac{1/(3n+2)}{1/n} = \lim_{n \to \infty}\frac{n}{3n+2} = \frac{1}{3}

The limit is finite and nonzero, so the series and the harmonic series converge or diverge together. The harmonic series diverges, so this one does too.

Direct comparison would be awkward here, because 13n+2<1n\frac{1}{3n+2} < \frac{1}{n} points the wrong way: being smaller than a divergent series proves nothing. Limit comparison sidesteps the direction problem entirely, which is why it is the right tool for terms that merely BEHAVE like a known series.

Constants never change the verdict

The terms are roughly 13n\frac{1}{3n}, which is one third of the harmonic terms. Multiplying every term of a divergent series by a positive constant leaves it divergent, because the partial sums are just scaled by that constant and still run away.

The same holds for convergence. Any series whose terms behave like cn\frac{c}{n} for a positive constant cc diverges, and any whose terms behave like cn2\frac{c}{n^{2}} converges. Once the shape is identified, the constant is irrelevant.

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

When should I use limit comparison instead of direct comparison?

When your series merely behaves like a known one rather than being cleanly bounded by it. Limit comparison needs no inequality, only a finite nonzero limit of the quotient.

Which series should I compare against?

Keep the highest-degree term on top and bottom and discard the rest. Here 13n+2\frac{1}{3n+2} reduces to 13n\frac{1}{3n}, so compare against 1/n\sum 1/n.