Q. 19 - Calculus

Question

Find two divergent geometric series $\sum _{k=0}^{\infty }{a}_k$ and $\sum _{k=0}^{\infty }{b}_k$ with all positive terms such that $\sum _{k=0}^{\infty }\frac{a_k}{b_k}$ converges.

Step-by-Step Solution

Answer

Ans: 

part (a). The divergent geometric series $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }{2}^k$

part (b). The divergent geometric series $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }{4}^k$

part (c). The series  is converge $\sum _{k=0}^{\infty }\frac{a_k}{b_k}=\sum _{k=0}^{\infty }\frac{1}{2^k}$

Step 1. Given Information:

Consider the two divergent geometric series $\sum _{k=0}^{\infty }{a}_k$ and $\sum _{k=0}^{\infty }{b}_k$ with all positive terms such that $\sum _{k=0 }^{\infty }\frac{a_{k }}{b_k}$converge.

Step 2. Finding the divergent geometric series ∑ k = 0 ∞ a k

Consider the geometric series $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }{2}^k$.

The series $\sum _{k=0}^{\infty }{a}_k$ is geometric series with common ratio $r=2$, which is greater than 1 . The geometric series with ratio greater than 1 is divergent.

Therefore, $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }{2}^k$is divergent.

Step 3. Finding the divergent geometric series ∑ k = 0 ∞ b k

Consider the geometric series $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }{4}^k$.

The series $\sum _{k=0}^{\infty }{4}^k$is geometric series with common ratio $r=4$, which is greater than 1 . The geometric series with ratio greater than 1 is divergent.

Therefore, $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }{4}^k$ is dlvergent.

Step 4. Finding the converges geometric series of ∑ k = 0   ∞ a k   b k with all positive terms:

The series $\sum _{k=0}^{\infty }\frac{a_k}{b_k}$ is

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{a_k}{b_k}=\sum _{k=0}^{\infty }\frac{2^k}{4^k} \\ =\sum _{k=0}^{\infty }\frac{1}{2^k} \end{gathered}$$

The series $\sum _{k=0}^{\infty }\frac{1}{2^x}$ 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, $\sum _{k=0}^{\infty }\frac{a_k}{b_k}=\sum _{k=0}^{\infty }\frac{1}{2^k}$ is convergent.