AP Calculus BC

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

The sum of 1 over n squared plus 1 converges. Its terms are smaller than 1 over n squared, which is a convergent p-series, so direct comparison proves convergence. The limit comparison test gives the same answer with a ratio of 1.

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

Converges

Settled by the limit comparison test.

Two comparisons, both clean

A bigger denominator means a smaller term, and the smaller series is dominated by one that already converges.

1n2+1<1n2for every n≥1\frac{1}{n^{2}+1} < \frac{1}{n^{2}} \quad \text{for every } n \ge 1

Direct comparison therefore gives convergence. The limit comparison test is the more forgiving version, and it confirms the two series behave identically.

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

When to prefer limit comparison

Direct comparison needs the inequality to point the right way, which can fail on a term like 1/(n^2 - 1). Limit comparison only needs a finite positive ratio, so it works whenever the two series have the same growth order.

The mistakes students make

  • Comparing in the wrong direction. To prove CONVERGENCE the terms must be smaller than a convergent series; being smaller than a divergent one proves nothing.
  • Splitting 1n2+1\frac{1}{n^{2}+1} into 1n2+11\frac{1}{n^{2}} + \frac{1}{1}. There is no such rule for a sum in a denominator.

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 comparison with the convergent pp-series ∑1n2\sum \frac{1}{n^{2}}.

Direct or limit comparison?

Either works here. Direct is available because the inequality points the right way; limit comparison gives a ratio of 11.