AP Calculus BC

Does the Sum of 1/n^(1/3) Converge? No

Q: Is the cube root of n a p-series?

Yes, with p = 1/3. Radicals in the denominator are fractional powers, and the test reads the exponent regardless of how it is written.

The series diverges. It is a p-series with p = 1/3, which is less than one, so it fails the threshold. The terms do tend to zero, which is why the nth term test is silent here and the exponent has to be read instead.

\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}

Diverges

Settled by the p-series test.

Vanishing terms are not enough

The terms $1/\sqrt[3]{n}$ certainly tend to 0, so the nth term test returns nothing. That is the standard trap: it proves divergence only when the terms fail to vanish, and it can never prove convergence.

These terms shrink more slowly than the harmonic terms $1/n$ , and the harmonic series already diverges. So this one has no chance, and the p-series test says so in one step.

How slowly it diverges

Divergence here is genuinely sluggish. The partial sums grow like $\frac{3}{2}n^{2/3}$ , so reaching a total of 100 takes around a quarter of a million terms.

That is exactly why numerical evidence is useless for this question. Summing terms on a computer would show a total creeping upward and looking like it might settle, and it never does. The exponent is the evidence, not the sums.

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

The terms go to zero, so why does it diverge?

Vanishing terms are necessary for convergence but nowhere near sufficient. The harmonic series is the standard example, and every p-series with $p \le 1$ behaves the same way.

Is the cube root of n a p-series?

Yes, with $p = 1/3$ . Radicals in the denominator are fractional powers, and the test reads the exponent regardless of how it is written.