AP Calculus BC

Does the Sum of 1/n^(2/3) Converge? No

The sum of 1 over n to the two thirds diverges. It is a p-series with p equal to two thirds, and p at or below 1 means divergence. The terms do go to zero, but more slowly than the harmonic terms do, and going to zero was never enough on its own.

∑n=1∞1n2/3\sum_{n=1}^{\infty}\frac{1}{n^{2/3}}

Diverges

Settled by the p-series test.

The radical is a power in disguise

Exams usually print this one as a cube root. A cube root is a one third power, so move the root into the exponent before deciding anything.

1n23=1n2/3,p=23\frac{1}{\sqrt[3]{n^{2}}} = \frac{1}{n^{2/3}}, \qquad p = \frac{2}{3}

With p=23<1p = \frac{2}{3} < 1, the p-series test returns divergence. Fractional exponents are perfectly legal here; the rule cares only which side of 11 the number lands on.

The terms shrink and it diverges anyway

At n=1000n = 1000 the term is 1100=0.01\frac{1}{100} = 0.01, while the harmonic term is 0.0010.001. These terms are ten times larger at that point, and at least as large as the harmonic terms everywhere, strictly larger from n=2n = 2 on.

Since ∑1n\sum \frac{1}{n} diverges, direct comparison forces the same verdict. That is a useful second opinion when you are unsure you read the exponent correctly.

Shrinking is necessary, not sufficient

Terms tending to zero is required for convergence, so failing it proves divergence outright. Passing it proves nothing. This series passes and diverges, which is exactly why the nth term test comes back inconclusive.

The mistakes students make

Almost every error on this series happens before any test is applied.

  • Taking p=3p = 3 from the index of the root, or p=2p = 2 from the exponent inside it. The value of pp is the quotient 23\frac{2}{3}.
  • Concluding convergence because the terms approach 00. That observation only tells you the nth term test is silent.
  • Assuming a p-series needs a whole number exponent, so the rule must not apply. Any real pp works, and 23\frac{2}{3} is below the cutoff.

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^(2/3) converge?

No. It is a p-series with p=23p = \frac{2}{3}, and p≤1p \le 1 means divergence.

Is 1/n^(2/3) the same as 1 over the cube root of n squared?

Yes. n23=n2/3\sqrt[3]{n^{2}} = n^{2/3}, and converting to the power form is what makes pp readable.

Why does it diverge if the terms go to zero?

Because they go to zero too slowly. They never drop below 1n\frac{1}{n}, and the harmonic series already diverges.