AP Calculus BC

Does the Sum of 1/n! Converge? Yes

The sum of 1 over n factorial converges. The ratio test gives a limit of 0, which is less than 1, so it converges and does so very fast. Starting from n equals 1 the sum is e minus 1; starting from n equals 0 it is e itself.

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

Converges

sum=e−1\text{sum} = e - 1

Settled by the ratio test.

Factorials are the ratio test's home ground

A factorial almost always signals the ratio test, because (n+1)!n!=n+1\frac{(n+1)!}{n!} = n+1 collapses in a single step.

an+1an=1(n+1)!⋅n!1=1n+1\frac{a_{n+1}}{a_{n}} = \frac{1}{(n+1)!}\cdot\frac{n!}{1} = \frac{1}{n+1}
L=lim⁡n→∞1n+1=0<1L = \lim_{n \to \infty}\frac{1}{n+1} = 0 < 1

A ratio limit of 00 is the strongest possible outcome: the terms shrink faster than any geometric series.

The series that defines e

This is the Maclaurin series for exe^{x} evaluated at x=1x = 1, with the n=0n = 0 term separated out.

ex=∑n=0∞xnn!  ⟹  e=∑n=0∞1n!=1+∑n=1∞1n!e^{x} = \sum_{n=0}^{\infty}\frac{x^{n}}{n!} \implies e = \sum_{n=0}^{\infty}\frac{1}{n!} = 1 + \sum_{n=1}^{\infty}\frac{1}{n!}

It converges extraordinarily fast

Ten terms already agree with e to seven decimal places. Factorial growth in the denominator outruns everything on the AP syllabus.

The mistakes students make

  • Reading (n+1)!n!\frac{(n+1)!}{n!} as n!n! or as (n+1)!(n+1)!. It is exactly n+1n+1, and that cancellation is the whole reason the test works here.
  • Reporting the sum as ee when the sum starts at n=1n = 1. The n=0n = 0 term contributes the missing 11.
  • Trying the pp-series rule. A factorial is not a power of nn.

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! converge?

Yes. The ratio test gives L=0<1L = 0 < 1.

What is its sum?

e−1e - 1 from n=1n = 1, or ee from n=0n = 0.

Why is the ratio test the right tool?

Because (n+1)!n!=n+1\frac{(n+1)!}{n!} = n+1, so the factorials cancel completely in one step.