91Ó°ÊÓ

Consider the two divergent geometric series $\underset{k=0}{\overset{∞}{∑}}{a}_k$and $\underset{k=0}{\overset{∞}{∑}}{b}_k$with all positive terms such that $\underset{k=0}{\overset{∞}{∑}}\frac{a_k}{b_k}$converge.

Consider the geometric series $\underset{k=0}{\overset{∞}{∑}}{a}_k=\underset{k=0}{\overset{∞}{∑}}{2}^k$.

The series $\underset{k=0}{\overset{∞}{∑}}{a}_k$is geometric series with common ratio $r=2$, which is greater than 1 . The geometric series with ratio greater than 1 is divergent.

Therefore, $\underset{k=0}{\overset{∞}{∑}}{a}_k=\underset{k=0}{\overset{∞}{∑}}{2}^k$is divergent.

Consider the geometric series $\underset{k=0}{\overset{∞}{∑}}{b}_k=\underset{k=0}{\overset{∞}{∑}}{4}^k$.

The series $\underset{k=0}{\overset{∞}{∑}}{4}^k$is geometric series with common ratio $r=4$, which is greater than 1 . The geometric series with ratio greater than 1 is divergent.

Therefore, $\underset{k=0}{\overset{∞}{∑}}{b}_k=\underset{k=0}{\overset{∞}{∑}}{4}^k$is dlvergent.

The series $\underset{k=0}{\overset{∞}{∑}}\frac{a_k}{b_k}$ is

$$\begin{gathered} \underset{k=0}{\overset{∞}{∑}}\frac{a_k}{b_k}=\underset{k=0}{\overset{∞}{∑}}\frac{2^k}{4^k} \\ =\underset{k=0}{\overset{∞}{∑}}\frac{1}{2^k} \end{gathered}$$

The series $\underset{k=0}{\overset{∞}{∑}}\frac{1}{2^x}$is a geometric series with common ratio $r=\frac{1}{2}$, which is less than 1 . The geometric series with ratio less than 1 is convergent.

Therefore, $\underset{k=0}{\overset{∞}{∑}}\frac{a_k}{b_k}=\underset{k=0}{\overset{∞}{∑}}\frac{1}{2^k}$ is convergent.