AP Calculus BC

Does the Sum of 1/n^(5/2) Converge? Yes

Q: Is 1 over n squared root n a p-series?

Yes. n^{2}\sqrt{n} = n^{5/2}, so p = 5/2 and the series converges. Convert radicals to fractional exponents before picking a test.

The series converges. It is a p-series with p = 5/2, which is greater than one, so the p-series test settles it immediately. A fractional exponent is treated exactly like a whole one; only its size against the threshold of one matters.

\sum_{n=1}^{\infty}\frac{1}{n^{5/2}}

Converges

Settled by the p-series test.

Fractional exponents are still p-series

Written with a radical the terms are $\frac{1}{n^{2}\sqrt{n}}$ , which can look like something needing a comparison. It is not: $n^{2}\sqrt{n} = n^{5/2}$ , and the p-series test reads $p = 5/2$ straight off.

Rewriting radicals as fractional powers before choosing a test is worth doing every time. A large share of comparison arguments on the exam are p-series wearing a square root.

Where the threshold sits

The only question is whether $5/2 > 1$ , and it plainly is. Compare with $\sum 1/\sqrt{n}$ , which has $p = 1/2 < 1$ and diverges, and with $\sum 1/n^{3/2}$ , which has $p = 3/2 > 1$ and converges.

All three have terms tending to 0, which is why the nth term test cannot separate them and the exponent has to be read.

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

Is 1 over n squared root n a p-series?

Yes. $n^{2}\sqrt{n} = n^{5/2}$ , so $p = 5/2$ and the series converges. Convert radicals to fractional exponents before picking a test.

Does the p-series test need p to be an integer?

No. Any real p works, including fractional and irrational exponents. The threshold is $p > 1$ and nothing else.