AP Calculus BC

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

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

Yes, absolutely, by the ratio test with L = \frac{1}{3}.

Q: What is the ratio limit for n^p over b^n?

It is \frac{1}{b} for every fixed p, because the polynomial part contributes a factor tending to 1.

Q: Would the root test work here as well?

Yes, and it gives the same L = \frac{1}{3}, since n^{1/n} \to 1.

Consecutive terms of n cubed over 3 to the n sit in the ratio one third times the cube of n plus 1 over n. The cube tends to 1, leaving a ratio limit of one third, and the base alone fixed that. Any limit below 1 means the sum converges absolutely.

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

Converges

Settled by the ratio test.

The ratio test in three lines

Write $\frac{a_{n+1}}{a_{n}}$ and let the powers of $3$ cancel against each other.

\frac{a_{n+1}}{a_{n}} = \frac{(n+1)^{3}}{3^{n+1}}\cdot\frac{3^{n}}{n^{3}} = \frac{1}{3}\left(\frac{n+1}{n}\right)^{3}

L = \lim_{n \to \infty}\frac{1}{3}\left(1 + \frac{1}{n}\right)^{3} = \frac{1}{3}

$L = \frac{1}{3}$ is below $1$ , so the series converges absolutely.

The base sets the limit, the power does not

Nothing above depended on the exponent $3$ on the numerator. For a fixed power $p$ and a base $b > 1$ the same cancellation runs through.

\lim_{n \to \infty}\frac{(n+1)^{p}}{b^{n+1}}\cdot\frac{b^{n}}{n^{p}} = \frac{1}{b}\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{p} = \frac{1}{b}

So $\sum \frac{n^{p}}{b^{n}}$ converges for every $b > 1$ , whatever $p$ happens to be. Raising $p$ makes the terms bigger for longer before they fall, and never touches the verdict.

The root test agrees

Taking nth roots gives n to the power 3 over n, divided by 3. Since n to the power 1 over n tends to 1, the root limit is also 1 over 3. Both arguments are valid and both give 1 over 3, but the ratio test is the one the CED assesses, so it is the one to write on the exam.

The mistakes students make

Watch the exponential factor, since that is where the answer lives.

Cancelling $3^{n+1}$ against $3^{n}$ as though they were equal. What is left over is exactly $\frac{1}{3}$ , and that leftover is the whole answer.
Expecting the cube to matter. $\left(\frac{n+1}{n}\right)^{3}$ tends to $1$ for any exponent at all.
Reporting $L = 0$ by dropping the numerator $(n+1)^{3}$ while keeping the $n^{3}$ below the line, which leaves $\frac{1}{3n^{3}} \to 0$ . The two polynomials cancel to a factor tending to $1$ , so neither may be discarded on its own.

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 n^3/3^n converge?

Yes, absolutely, by the ratio test with $L = \frac{1}{3}$ .

What is the ratio limit for n^p over b^n?

It is $\frac{1}{b}$ for every fixed $p$ , because the polynomial part contributes a factor tending to $1$ .

Would the root test work here as well?

Yes, and it gives the same $L = \frac{1}{3}$ , since $n^{1/n} \to 1$ .