AP Calculus BC

Does the Sum of 3^n/4^n Converge? Yes, to 3

Q: How do I see it is geometric?

Combine the equal exponents: \frac{3^{n}}{4^{n}} = \left(\frac{3}{4}\right)^{n}, a constant ratio of \frac{3}{4}.

The sum of 3 to the n over 4 to the n, starting at n equals 1, converges to exactly 3. Combining the powers turns the term into three quarters raised to the n, so it is geometric with ratio three quarters, which is less than 1.

\sum_{n=1}^{\infty}\frac{3^{n}}{4^{n}}

Converges

\text{sum} = 3

Settled by the geometric series test.

Combine the powers first

Two powers with the same exponent collapse into one, which is what exposes the ratio.

\frac{3^{n}}{4^{n}} = \left(\frac{3}{4}\right)^{n}

Now $a = \frac{3}{4}$ and $r = \frac{3}{4}$ , and the sum formula finishes it.

\sum_{n=1}^{\infty}\left(\frac{3}{4}\right)^{n} = \frac{\frac{3}{4}}{1-\frac{3}{4}} = \frac{\frac{3}{4}}{\frac{1}{4}} = 3

A ratio near 1 converges slowly

With r = 3/4 it takes about 30 terms to get within 0.001 of 3, against 10 for r = 1/2. Convergence is convergence, but the rate is governed entirely by how close the ratio sits to 1.

The mistakes students make

Treating the numerator and denominator as separate series. $\sum 3^{n}$ and $\sum 4^{n}$ both diverge; the quotient is what converges.
Reading the ratio as $\frac{4}{3}$ , which is greater than $1$ and would mean divergence. The base of the numerator goes on top.

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/4^n converge?

Yes, to $3$ starting from $n = 1$ .

How do I see it is geometric?

Combine the equal exponents: $\frac{3^{n}}{4^{n}} = \left(\frac{3}{4}\right)^{n}$ , a constant ratio of $\frac{3}{4}$ .