AP Calculus BC

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

The sum of 1 over n cubed converges. It is the p-series with p = 3, comfortably greater than 1. Unlike the p equals 2 case there is no simple closed form for the sum: the value is about 1.2021, known as Apery's constant.

∑n=1∞1n3\sum_{n=1}^{\infty}\frac{1}{n^{3}}

Converges

Settled by the p-series test.

Straight from the p-series rule

With p=3>1p = 3 > 1 the series converges, and the integral test confirms the bound.

∫1∞dxx3=[−12x2]1∞=12\int_{1}^{\infty}\frac{dx}{x^{3}} = \left[-\frac{1}{2x^{2}}\right]_{1}^{\infty} = \frac{1}{2}

The larger the exponent, the faster the terms decay and the more quickly the partial sums settle. This one is within 0.0010.001 of its limit after about twenty terms.

Convergent does not mean summable

The value here, about 1.2021, has no closed form in terms of pi or elementary constants. It was not even proved irrational until 1978. Knowing a series converges says nothing about being able to name its sum.

The mistakes students make

  • Expecting a π\pi-based value by analogy with ∑1n2=π26\sum \frac{1}{n^{2}} = \frac{\pi^{2}}{6}. The odd exponents behave completely differently.
  • Using the ratio test, which returns L=1L = 1 and is inconclusive for every pp-series.

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 converge?

Yes, by the pp-series test with p=3>1p = 3 > 1.

What is its sum?

About 1.20211.2021, called Apery's constant. There is no elementary closed form, unlike the p=2p = 2 case.