Q. 19 - Calculus

Question

Give an example of 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.

Step-by-Step Solution

Answer

 The series$\sum _{k=1}^{\infty }{a}_k{b}_k=\sum _{k=1}^{\infty }\left(\frac{1}{k^2}\right)$ is convergent.

Step 1. Given information

$\sum _{k=1}^{\infty }{a}_k$ and $\sum _{k=1}^{\infty }{b}_k$ are divergent.

Step 2. Check if ∑ a k k = 1 ∞ converges

Calculating $\frac{a_{k+1}}{a_k}$, 

$a_k=\frac{1}{k}$

$a_{k+1}=\frac{1}{(k+1)}$

$\begin{array}{r}\frac{a_{k+1}}{a_k}=\frac{\frac{1}{(k+1)}}{\frac{1}{k}}\\ =\frac{k}{(k+1)}\end{array}$

Now, taking limits,

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} \frac{a_{k+1}}{a_k}=\underset{k\to \mathrm{\infty }}{lim} \frac{k}{(k+1)}\\ =\left(\underset{k\to \mathrm{\infty }}{lim} \frac{k}{(k+1)}\right)\\ =\mathrm{\infty }\end{array}$

Hence, the series $\underset{k=1}{\overset{\infty }{\sum a_k}}=\sum _{k=1}^{\infty }\frac{1}{k}$diverges.

Similarly, $\underset{k=1}{\overset{\infty }{\sum b_k}}=\sum _{k=1}^{\infty }\frac{1}{k}$ isdivergent.

Step 3. Check if ∑ a k b k k = 1 ∞ is divergent or convergent

$\begin{array}{r}\sum _{k=1}^{\mathrm{\infty }} a_k{b}_k=\sum _{k=1}^{\mathrm{\infty }} \left(\frac{1}{k}\right)\left(\frac{1}{k}\right)\\ =\sum _{k=1}^{\mathrm{\infty }} \left(\frac{1}{k^2}\right)\end{array}$

Hence, $\sum _{k=1}^{\mathrm{\infty }} a_k{b}_k$ is convergent.