AP Calculus BC

Does the Sum of 2^n/5^n Converge? Yes, to 2/3

Q: Does the sum of 2^n/5^n converge?

Yes, to \frac{2}{3} starting from n = 1.

Q: How do I see the ratio?

Combine the equal exponents: \frac{2^{n}}{5^{n}} = \left(\frac{2}{5}\right)^{n}.

The sum of 2 to the n over 5 to the n, starting at n equals 1, converges to two thirds. Combining the equal exponents turns the term into two fifths raised to the n, a geometric series with ratio two fifths.

\sum_{n=1}^{\infty}\frac{2^{n}}{5^{n}}

Converges

\text{sum} = \frac{2}{3}

Settled by the geometric series test.

Combine, then apply the formula

\frac{2^{n}}{5^{n}} = \left(\frac{2}{5}\right)^{n}

\sum_{n=1}^{\infty}\left(\frac{2}{5}\right)^{n} = \frac{\frac{2}{5}}{1-\frac{2}{5}} = \frac{\frac{2}{5}}{\frac{3}{5}} = \frac{2}{3}

A ratio of $\frac{2}{5}$ is well below $1$ , so this converges quickly: twelve terms put you within $0.0001$ of the total.

The mistakes students make

Treating numerator and denominator as separate series. Both $\sum 2^{n}$ and $\sum 5^{n}$ diverge; only the quotient converges.
Inverting the ratio to $\frac{5}{2}$ , which exceeds $1$ and would wrongly suggest divergence.

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/5^n converge?

Yes, to $\frac{2}{3}$ starting from $n = 1$ .

How do I see the ratio?

Combine the equal exponents: $\frac{2^{n}}{5^{n}} = \left(\frac{2}{5}\right)^{n}$ .