AP Calculus BC

Does the Sum of (-1)^n/n! Converge? Yes, to 1/e - 1

The sum of negative 1 to the n over n factorial converges ABSOLUTELY, to 1 over e minus 1. The alternating series test applies, since 1 over n factorial decreases to 0. The ratio test on the absolute values gives a limit of 0, which upgrades the verdict to absolute.

∑n=1∞(−1)nn!\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}

Converges

sum=1e−1\text{sum} = \frac{1}{e}-1

Settled by the alternating series test.

The alternating series test, in two checks

The magnitudes are bn=1n!b_{n} = \frac{1}{n!}. Each one is a fraction of the one before it, so they strictly decrease.

bn+1=1(n+1)!=1n+1⋅1n!<bnb_{n+1} = \frac{1}{(n+1)!} = \frac{1}{n+1}\cdot\frac{1}{n!} < b_{n}

Factorials outgrow everything, so bn→0b_{n} \to 0. Both conditions hold, and the series converges.

Absolute convergence is the stronger statement

The alternating series test says the signed series converges. It says nothing at all about ∑1n!\sum \frac{1}{n!}, so that has to be handled separately. The ratio test is quick.

lim⁡n→∞∣an+1an∣=lim⁡n→∞n!(n+1)!=lim⁡n→∞1n+1=0\lim_{n \to \infty}\left|\frac{a_{n+1}}{a_{n}}\right| = \lim_{n \to \infty}\frac{n!}{(n+1)!} = \lim_{n \to \infty}\frac{1}{n+1} = 0

A limit below 11 means ∑1n!\sum \frac{1}{n!} converges, so the original series converges absolutely. It is the limit that decides, not the individual ratios, and the ratio test says nothing at all when that limit is 11. Absolute convergence is worth more than plain convergence: an absolutely convergent series keeps its sum under any rearrangement, and it can be added to and multiplied by other series safely.

Where 1/e minus 1 comes from

This is the exponential series evaluated at x=−1x = -1.

ex=∑n=0∞xnn!⟹e−1=∑n=0∞(−1)nn!e^{x} = \sum_{n=0}^{\infty}\frac{x^{n}}{n!} \quad\Longrightarrow\quad e^{-1} = \sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}

That version starts at n=0n = 0, where the term is 11. This series starts at n=1n = 1, so the missing term has to come off.

∑n=1∞(−1)nn!=1e−1≈−0.6321\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!} = \frac{1}{e} - 1 \approx -0.6321

The mistakes students make

The index and the word absolute are where marks go missing here.

  • Quoting the sum as 1e\frac{1}{e}. The exponential series starts at n=0n = 0, and dropping that term of 11 leaves 1e−1\frac{1}{e} - 1.
  • Calling the convergence conditional because the series alternates. Alternating and conditional are separate ideas, and ∑∣an∣\sum |a_{n}| decides which applies.
  • Believing the alternating series test settled absolute convergence. It never can, because it only ever looks at the signed 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/n! converge?

Yes, absolutely, and the value is 1e−1≈−0.6321\frac{1}{e} - 1 \approx -0.6321.

Is it absolutely or conditionally convergent?

Absolutely. The ratio test on ∑1n!\sum \frac{1}{n!} gives a limit of 00, so the absolute series converges too.

Why is the sum not 1/e?

Because the familiar identity runs from n=0n = 0 and this series runs from n=1n = 1. The n=0n = 0 term is 11, so the value drops by 11.