AP Calculus BC

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

Q: How do I compare degrees when a square root is involved?

Halve the degree under the root: \sqrt{n^{2}+1} counts as degree 1, and \sqrt{n^{3}} as degree 3/2. Then compare with the numerator as usual.

The series diverges. Top and bottom grow at the same rate, so the terms tend to one rather than zero and the nth term test settles it. The square root disguises the degree comparison but does not change it.

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

Diverges

Settled by the nth term test for divergence.

Compare the degrees through the root

A square root halves the degree: $\sqrt{n^{2}+1}$ behaves like $n$ for large $n$ . So the numerator and denominator are both degree one, and the quotient tends to the ratio of leading coefficients, which is 1.

\frac{n}{\sqrt{n^{2}+1}} = \frac{1}{\sqrt{1 + 1/n^{2}}} \longrightarrow 1

Dividing top and bottom by $n$ is the reliable move, and inside the root it becomes dividing by $n^{2}$ . Getting that step wrong is the most common source of a wrong limit here.

Contrast with the convergent neighbour

The series $\sum \frac{n}{n^{2}+1}$ , without the root, has terms behaving like $1/n$ and also diverges, but for a different reason: there the terms DO tend to 0 and the nth term test is silent, so limit comparison with the harmonic series is needed.

Two similar-looking series, two different tests. Which test applies is decided by the limit of the terms, so computing that limit is always the first move.

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

How do I compare degrees when a square root is involved?

Halve the degree under the root: $\sqrt{n^{2}+1}$ counts as degree 1, and $\sqrt{n^{3}}$ as degree $3/2$ . Then compare with the numerator as usual.

Why is this different from n over n squared plus one?

Without the root, the denominator is degree 2 against a degree 1 numerator, so the terms behave like $1/n$ and tend to 0. The nth term test then says nothing and a comparison is needed.