AP Calculus BC

Does the Sum of n^n/n! Converge? No

The sum of n to the n over n factorial diverges. The ratio test gives a limit of e, about 2.718, which is greater than 1. It is the exact reciprocal of the n factorial over n to the n series, and the reciprocal ratio flips the verdict.

∑n=1∞nnn!\sum_{n=1}^{\infty}\frac{n^{n}}{n!}

Diverges

Settled by the ratio test.

The reciprocal series, the reciprocal ratio

an+1an=(n+1)n+1(n+1)!⋅n!nn=(1+1n)n⟶e\frac{a_{n+1}}{a_{n}} = \frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^{n}} = \left(1+\frac{1}{n}\right)^{n} \longrightarrow e

Since e≈2.718>1e \approx 2.718 > 1, the ratio test proves divergence. Each term is eventually about 2.72.7 times the one before, so the terms grow without bound.

Both directions are decisive

A ratio limit above 1 proves divergence just as firmly as a limit below 1 proves convergence. Only L exactly 1 leaves the test with nothing to say.

The nth term test also works

Since the terms grow without bound they certainly do not tend to 00, so the nth term test settles it in one line. When two tests apply, the shorter justification is the one to write.

The mistakes students make

  • Assuming a factorial denominator always wins. Against nnn^{n} it loses.
  • Reading L=eL = e as inconclusive. Only L=1L = 1 is inconclusive; ee is comfortably above it.

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 n^n/n! converge?

No. The ratio test gives L=e>1L = e > 1, so it diverges.

How does it relate to n!/n^n?

It is the reciprocal series, and its ratio limit is the reciprocal, ee instead of 1e\frac{1}{e}, which flips the verdict.