AP Calculus BC

Does the Sum of 1/n^(3/2) Converge? Yes

Q: Does the sum of 1/n^(3/2) converge?

Yes, by the p-series test with p = 1.5 > 1.

Q: Can p be a fraction?

Yes. The rule is simply p > 1, and p may be any real number.

The sum of 1 over n to the three halves converges. It is the p-series with p equal to 1.5, and since that is greater than 1 the series converges. It is worth knowing because it shows p does not need to be a whole number, only greater than 1.

\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}

Converges

Settled by the p-series test.

p does not have to be an integer

The rule is a strict inequality on a real number, not a statement about whole numbers. Here $p = \frac{3}{2} = 1.5 > 1$ , so the series converges.

\frac{1}{n^{3/2}} = \frac{1}{n\sqrt{n}}

That second form is how the term usually appears on an exam, disguised as a product rather than a single power. Recognising $n\sqrt{n}$ as $n^{3/2}$ is the whole task.

The boundary is genuinely sharp

p = 1.01 converges and p = 0.99 diverges, and no numerical experiment can tell them apart: after ten million terms both partial sums are still climbing at about two per decade. The theorem is doing work no computation could.

The mistakes students make

Reading $n\sqrt{n}$ as $n^{2}$ or as $\sqrt{n^{2}}$ . It is $n^{1} \cdot n^{1/2} = n^{3/2}$ .
Thinking $p$ must be a whole number, and so forcing the term into the wrong comparison.

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^(3/2) converge?

Yes, by the $p$ -series test with $p = 1.5 > 1$ .

Can p be a fraction?

Yes. The rule is simply $p > 1$ , and $p$ may be any real number.