AP Calculus BC

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

The series diverges. The denominator exceeds the numerator by only one degree, so the terms behave like one over n, and limit comparison with the harmonic series gives a limit of one. A gap of one degree is never enough.

∑n=1∞n2+1n3+1\sum_{n=1}^{\infty}\frac{n^{2}+1}{n^{3}+1}

Diverges

Settled by the limit comparison test.

The degree gap is the whole story

For a quotient of polynomials, subtract the degrees: 3−2=13 - 2 = 1, so the terms behave like 1/n11/n^{1}. The p-series threshold is p>1p > 1 STRICTLY, so a gap of exactly one lands on the divergent side.

lim⁡n→∞(n2+1)/(n3+1)1/n=lim⁡n→∞n3+nn3+1=1\lim_{n \to \infty}\frac{(n^{2}+1)/(n^{3}+1)}{1/n} = \lim_{n \to \infty}\frac{n^{3}+n}{n^{3}+1} = 1

A finite nonzero limit, so the series shares the harmonic series' fate and diverges.

The rule this gives you

For any quotient of polynomials, the series converges exactly when the degree of the denominator exceeds the degree of the numerator by MORE than one. A gap of 2 gives p=2p = 2 and convergence; a gap of 1 gives the harmonic case and divergence.

The terms here do tend to 0, so the nth term test is silent. That is the usual situation, and it is why the degree gap rather than the vanishing of the terms is what to look at first for rational 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

What degree gap is needed for convergence?

Strictly more than one. A gap of exactly one reproduces the harmonic series and diverges; a gap of 1.51.5 or 2 converges. Fractional gaps from radicals count too.

Do the constant terms matter?

No. The plus ones affect the sum but not the verdict, since the limit comparison sends them to nothing. Only the leading degrees decide.