Q. 16 - Calculus

Question

Find two divergent series $\sum _{k=1}^{\infty }{a}_k$ and $\sum _{k=1}^{\infty }{b}_k$ such that the series $\sum _{k=1}^{\infty }(a_k-b_k)$ converges. Carefully use the sequence of partial sums for this new series to show that your answer is correct.

Step-by-Step Solution

Answer

Ans: 

part (a). divergent series of $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }1$

part (b). divergent series of $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\left(1\right)$

part (c).The partial sum of series  $=\sum _{k=0}^{\infty }0$ is a constant and hence, is convergent. Therefore, $\sum _{k=0}^{\infty }\left(a_k-b_k\right)=\sum _{k=0}^{\infty }0$ is convergent.

Step 1. Given information:

Consider the two divergent geometric series $\sum _{k=1}^{\infty }{a}_k$ and $\sum _{k=1}^{\infty }{b}_k$ such that $\sum _{k=1}^{\infty }\left(a_k-b_k\right)$converge.

Step 2. Finding divergent series of ∑ k = 1 ∞ a k :

Consider the geometric series $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }1$.

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

Therefore, $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }1$ is divergent.

Step 3. Finding divergent series of ∑ k = 1 ∞ b k :

Consider the geometric series $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\left(1\right)$.

The series $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\left(1\right)$ is a geometric series with common ratio $r=1$, which is equal to 1 .

The geometric series with ratio equal to 1 is divergent.

Therefore, $\sum _{k=0}^{\infty }{b}_k=\sum _{k=0}^{\infty }\left(1\right)$ is divergent.

Step 4. Using the sequence of partial sums in new series :

The series $\sum _{k=0}^{\infty }\left(a_k-b_k\right)$ is

$$\begin{gathered} \sum _{k=0}^{\infty }\left(a_k-b_k\right)=\sum _{k=0}^{\infty }\left(1-1\right) \\ \sum _{k=0}^{\infty }\left(0\right)=0 \end{gathered}$$

The partial sum of series $=\sum _{k=0}^{\infty }0$ is a constant and hence, is convergent. Therefore, $\sum _{k=0}^{\infty }\left(a_k-b_k\right)=\sum _{k=0}^{\infty }0$ is convergent.