AP Calculus BC

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

The series converges. Keeping only the dominant powers, the terms behave like one over n to the three halves, and that p-series converges because the exponent exceeds one. Limit comparison makes the argument precise.

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

Converges

Settled by the limit comparison test.

Finding the comparison series

Keep only the dominant behaviour: $\sqrt{n+1}$ acts like $\sqrt{n} = n^{1/2}$ , so the terms act like $\frac{n^{1/2}}{n^{2}} = n^{-3/2}$ . That points at the p-series with $p = 3/2$ .

\lim_{n \to \infty}\frac{\sqrt{n+1}/n^{2}}{1/n^{3/2}} = \lim_{n \to \infty}\sqrt{\frac{n+1}{n}} = 1

A finite nonzero limit means the two series share a verdict, and $p = 3/2 > 1$ gives convergence.

Subtracting exponents carefully

The arithmetic to get right is $\frac{1}{2} - 2 = -\frac{3}{2}$ , so $p = \frac{3}{2}$ . Reading the exponent as $2$ , or as $\frac{1}{2}$ , is where this class of problem usually goes wrong, and both misreadings would still land on the correct verdict here, which is why the habit matters more than the outcome.

A worked contrast: $\frac{\sqrt{n+1}}{n}$ behaves like $n^{-1/2}$ , so $p = 1/2$ and it DIVERGES. Same numerator, one power less in the denominator, opposite answer.

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 pick the comparison series?

Take the highest power in the numerator over the highest in the denominator and subtract exponents. Whatever power of n survives is the p for the comparison.

Does the plus one under the root matter?

Not for the verdict. The limit comparison gives 1, so the shift is invisible to the classification. It affects the sum, not whether the sum exists.