AP Calculus BC

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

Q: Does the sum of 2^n/n! converge?

Yes, by the ratio test with L = 0.

Q: Would a bigger base change the answer?

No. For any constant k the ratio is \frac{k}{n+1} \to 0, so it always converges.

Q: What is the sum?

e^{2} - 1 from n = 1, about 6.389.

The sum of 2 to the n over n factorial converges. The ratio test gives 2 over n plus 1, which tends to 0, so it converges. Factorial growth outruns exponential growth, and from n equals 1 the sum is e squared minus 1.

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

Converges

\text{sum} = e^{2} - 1

Settled by the ratio test.

Factorial beats exponential

\frac{a_{n+1}}{a_{n}} = \frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^{n}} = \frac{2}{n+1}

L = \lim_{n \to \infty}\frac{2}{n+1} = 0 < 1

The numerator multiplies by a fixed $2$ each step while the denominator multiplies by an ever larger $n+1$ . Past $n = 2$ the denominator wins every single step, and it keeps winning by more.

The base never matters

Replace 2 with any constant k and the ratio becomes k/(n+1), still tending to 0. Every series of the form k^n over n factorial converges, no matter how large k is.

The sum comes from the exponential series

e^{2} = \sum_{n=0}^{\infty}\frac{2^{n}}{n!} = 1 + \sum_{n=1}^{\infty}\frac{2^{n}}{n!}

So the sum from $n = 1$ is $e^{2} - 1 \approx 6.389$ . As always, the convergence test gives the verdict and the series identity gives the value.

The mistakes students make

Assuming a large base forces divergence. $\sum \frac{100^{n}}{n!}$ converges too, just after a longer climb.
Inverting the ratio to $\frac{n+1}{2}$ and concluding divergence. The ratio is newer over older.

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 2^n/n! converge?

Yes, by the ratio test with $L = 0$ .

Would a bigger base change the answer?

No. For any constant $k$ the ratio is $\frac{k}{n+1} \to 0$ , so it always converges.

What is the sum?

$e^{2} - 1$ from $n = 1$ , about $6.389$ .