AP Calculus BC

Does the Sum of n/(n^3+1) Converge? Yes

The sum of n over n cubed plus 1 converges. The degree gap between denominator and numerator is 2, so the terms behave like 1 over n squared, a convergent p-series. Both direct and limit comparison work here.

∑n=1∞nn3+1\sum_{n=1}^{\infty}\frac{n}{n^{3}+1}

Converges

Settled by the limit comparison test.

Both comparisons are available

nn3+1<nn3=1n2\frac{n}{n^{3}+1} < \frac{n}{n^{3}} = \frac{1}{n^{2}}

The inequality points the right way, so direct comparison already proves convergence. Limit comparison agrees, with a ratio of 11.

lim⁡n→∞nn3+11n2=lim⁡n→∞n3n3+1=1\lim_{n \to \infty}\frac{\frac{n}{n^{3}+1}}{\frac{1}{n^{2}}} = \lim_{n \to \infty}\frac{n^{3}}{n^{3}+1} = 1

Adding to a denominator helps you

The +1 makes the terms smaller than n/n^3, which is exactly the direction direct comparison needs for a convergence proof. A MINUS in the denominator would flip that and force limit comparison instead.

The mistakes students make

  • Comparing with 1n3\frac{1}{n^{3}} and forgetting the numerator. The gap is 3−1=23 - 1 = 2.
  • Reporting divergence because the numerator grows. What matters is the gap, not the individual degrees.

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 n/(n^3+1) converge?

Yes. The degree gap is 22, so it behaves like ∑1n2\sum \frac{1}{n^{2}}.

Direct or limit comparison?

Either. The +1+1 makes the terms smaller than 1n2\frac{1}{n^{2}}, so direct comparison is available.