AP Calculus BC

Does the Sum of 3^n/n! Converge? Yes

Q: Does the sum of 3^n/n! converge?

Yes, by the ratio test with L = 0.

Q: Why do the terms grow at first?

The ratio \frac{3}{n+1} exceeds 1 until n = 2, so the terms rise to a peak at n = 2 and n = 3 before falling.

Q: What is the sum?

e^{3} - 1 from n = 1, about 19.086.

The sum of 3 to the n over n factorial converges. The ratio test gives 3 over n plus 1, which tends to 0. The terms actually GROW until n reaches 3 and only then start shrinking, which is a useful reminder that early behaviour says nothing about convergence.

\sum_{n=1}^{\infty}\frac{3^{n}}{n!}

Converges

\text{sum} = e^{3} - 1

Settled by the ratio test.

The ratio test, and where the terms turn

\frac{a_{n+1}}{a_{n}} = \frac{3}{n+1} \longrightarrow 0

The ratio exceeds $1$ while $n + 1 < 3$ , so the terms increase from $n = 1$ to $n = 3$ before turning over. The largest terms are $a_{2} = a_{3} = 4.5$ .

Convergence is about the tail

Only the eventual behaviour matters. A series can grow for a hundred terms and still converge, which is why the ratio test asks for a LIMIT rather than a comparison at any one n.

The mistakes students make

Concluding divergence from the first few terms increasing. Finitely many terms never decide convergence.
Checking $\frac{a_{n+1}}{a_{n}}$ at one value of $n$ instead of taking the limit.

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 3^n/n! converge?

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

Why do the terms grow at first?

The ratio $\frac{3}{n+1}$ exceeds $1$ until $n = 2$ , so the terms rise to a peak at $n = 2$ and $n = 3$ before falling.

What is the sum?

$e^{3} - 1$ from $n = 1$ , about $19.086$ .