AP Calculus BC

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

The series converges. It is a p-series with p = 5, and every p-series with p greater than one converges. Its exact value is a number known as zeta of 5, roughly 1.0369, which has no known closed form in terms of familiar constants.

∑n=1∞1n5\sum_{n=1}^{\infty}\frac{1}{n^{5}}

Converges

Settled by the p-series test.

One glance settles it

The terms are a pure power of nn in the denominator, so the p-series test applies directly: read off p=5p = 5, compare it with 1, and stop. No comparison, no integral, no ratio.

Convergence here is fast. By n=10n = 10 the terms are already down to 10−510^{-5}, so a handful of terms pins the sum to several decimal places, unlike the boundary cases near p=1p = 1.

Why no closed form is quoted

For even exponents the sums are famous: ∑1/n2=π2/6\sum 1/n^{2} = \pi^{2}/6 and ∑1/n4=π4/90\sum 1/n^{4} = \pi^{4}/90. For odd exponents past 1 no such formula is known. The value of ∑1/n5\sum 1/n^{5} is a genuine open question, not something left out for brevity.

This is worth knowing on the exam: a question can ask you to prove convergence without ever asking for the sum, and for most convergent series the sum is not available at all.

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

What is the exact sum of 1 over n to the fifth?

There is no known closed form. The value is ζ(5)≈1.036928\zeta(5) \approx 1.036928, and whether it can be written in terms of π\pi and elementary constants is still unsolved.

Does a larger p make convergence faster?

Yes. The larger p is, the faster the terms shrink and the sooner the partial sums settle. The verdict, though, only depends on whether p exceeds 1.