Question

Give an example of divergent series $\sum _{k=1}^{\infty }{a}_k$and $\sum _{k=1}^{\infty }{b}_k$such that the series role="math" localid="1649434191353" $\sum _{k=1}^{\infty }{a}_k{b}_k$converges.

Short answer

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

Step-by-step solution

Step 1. Given information

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

Step 2. Check if ∑akk=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 role="math" localid="1649434828493" $\underset{k=1}{\overset{\infty }{\sum a_k}}=\sum _{k=1}^{\infty }\frac{1}{k}$diverges.

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

Step 3. Check if ∑akbkk=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.