AP Calculus BC

Does the Sum of 5^n/4^n Converge? No

Q: Does the sum of 5^n/4^n converge?

No. It is geometric with r = \frac{5}{4} > 1, so it diverges.

Q: What is the sum of 5^n/4^n?

There is none. A divergent series has no sum, and \frac{a}{1-r} is not available once |r| \ge 1.

Q: Why does 3^n/4^n converge but 5^n/4^n not?

The ratios are \frac{3}{4} and \frac{5}{4}. One is below 1 and one is above, and that single comparison decides both.

The sum of 5 to the n over 4 to the n diverges. It is geometric with common ratio 5 over 4, which is above 1, so the geometric test returns divergence. The terms grow by 25 percent at every step, so they never approach zero and the nth term test settles it as well.

\sum_{n=1}^{\infty}\frac{5^{n}}{4^{n}}

Diverges

Settled by the geometric series test.

One power, ratio 5/4

Matching exponents on top and bottom mean the whole term is a single power, and the base of that power is the common ratio.

\frac{5^{n}}{4^{n}} = \left(\frac{5}{4}\right)^{n}, \qquad r = \frac{5}{4}

Since $|r| \ge 1$ , the geometric test returns divergence. Sitting only $0.25$ above the boundary buys nothing, because the rule has no near misses.

The terms grow, so the nth term test agrees

Each term is a quarter larger than the one before it, so the terms run away from $0$ instead of towards it.

\lim_{n\to\infty}\left(\frac{5}{4}\right)^{n} = \infty \ne 0

Set this beside $\sum_{n=1}^{\infty}\frac{3^{n}}{4^{n}}$ , where the ratio is $\frac{3}{4}$ and the series converges to $3$ . Which base is bigger is the whole story.

The mistakes students make

The tempting move is to reach for the sum formula regardless.

Applying $\frac{a}{1-r}$ anyway and reporting $-5$ . A series of positive terms cannot have a negative sum, and the formula is only valid when $|r| < 1$ .
Judging the ratio by how close the bases look. Bases $5$ and $4$ are near each other, but $r = \frac{5}{4}$ and any ratio at or above $1$ diverges.
Adding the first few terms, seeing modest numbers, and guessing convergence. The first term is $1.25$ and the tenth is already about $9.3$ , and in any case a handful of terms decides nothing.

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 5^n/4^n converge?

No. It is geometric with $r = \frac{5}{4} > 1$ , so it diverges.

What is the sum of 5^n/4^n?

There is none. A divergent series has no sum, and $\frac{a}{1-r}$ is not available once $|r| \ge 1$ .

Why does 3^n/4^n converge but 5^n/4^n not?

The ratios are $\frac{3}{4}$ and $\frac{5}{4}$ . One is below $1$ and one is above, and that single comparison decides both.