AP Calculus BC

Does the Sum of 2^n·n!/n^n Converge? Yes

Q: Why does e appear in a series with no exponentials in it?

The ratio produces \left(1+\frac{1}{n}\right)^{-n}, and that expression converging to 1/e is the definition of e. Any ratio mixing n! with n^{n} will surface it.

The series converges. The ratio test gives a limit of 2 divided by e, roughly 0.736, which is less than one. The constant e appears because the ratio contains the limit that defines it, and the verdict genuinely hinges on 2 being smaller than e.

\sum_{n=1}^{\infty}\frac{2^{n}\,n!}{n^{n}}

Converges

Settled by the ratio test.

Where e comes from

\frac{a_{n+1}}{a_n} = \frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{n}}{2^{n}\,n!} = 2\left(\frac{n}{n+1}\right)^{n}

The remaining bracket is $\left(1 + \frac{1}{n}\right)^{-n}$ , and $\left(1+\frac{1}{n}\right)^{n} \to e$ is the definition of $e$ . So the ratio tends to $2/e \approx 0.7358$ .

Since $2/e < 1$ , the series converges absolutely. This is one of the few AP-level series where the decisive constant is not a rational number.

How close the call is

The verdict rests on $2 < e \approx 2.71828$ . Replace the base 2 by 3 and the ratio becomes $3/e \approx 1.104 > 1$ , and the series diverges. Replace it by $e$ itself and the ratio is exactly 1, where the ratio test is INCONCLUSIVE and a finer argument is needed.

Stirling's approximation explains why: $n! \approx \sqrt{2\pi n}\,(n/e)^{n}$ , so the terms behave like $\sqrt{2\pi n}\,(2/e)^{n}$ , which is a geometric decay with exactly that ratio.

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

Why does e appear in a series with no exponentials in it?

The ratio produces $\left(1+\frac{1}{n}\right)^{-n}$ , and that expression converging to $1/e$ is the definition of e. Any ratio mixing $n!$ with $n^{n}$ will surface it.

What happens if the base is e exactly?

The ratio limit is exactly 1, where the ratio test says nothing. Deciding that case needs Stirling's approximation, which shows the terms behave like $\sqrt{2\pi n}$ and the series diverges.