Question

If\({\rm{\{ }}{{\rm{a}}_{\rm{n}}}{\rm{\} }}\)and \({\rm{\{ }}{{\rm{b}}_{\rm{n}}}{\rm{\} }}\)are divergent, then\({\rm{\{ }}{{\rm{a}}_{\rm{n}}}{{\rm{b}}_{\rm{n}}}{\rm{\} }}\)is divergent.

Short answer

Given statement is false; if\(\{ {a_n}\} \)and \(\{ {b_n}\} \)are divergent, then\(\{ {a_n}{b_n}\} \)is convergent.

Step-by-step solution

Convergence or Divergence of series

If the sequence\({{\rm{S}}_{\rm{n}}}\)is convergent and\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{S}}_{\rm{n}}}{\rm{ = s}}\)exists as a real number; then the series\(\sum {{{\rm{S}}_{\rm{n}}}} \)is convergent.

If\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{S}}_{\rm{n}}}\)does not exist or is plus or minus infinity, then the series\(\sum {{{\rm{S}}_{\rm{n}}}} \)is divergent.

Applying convergence or divergence test

Given statement is\(\{ {a_n}\} \)and \(\{ {b_n}\} \)are divergent.

Suppose, \({a_n} = {( - 1)^n}\) and \({b_n} = {( - 1)^{n + 1}}\).

That means value of \({a_n}\)is \( - 1\) for all odd \(n\)and \({b_n}\)is \( - 1\) for all even \(n\).

If we multiply both,

\(\begin{array}{c}{a_n} \cdot {b_n} = {( - 1)^n} \cdot {( - 1)^{n + 1}}\\ = {( - 1)^{2n + 1}}\\ = - 1\end{array}\)

As \(\{ {a_n}\} \)and \(\{ {b_n}\} \)are divergent but \(\{ {a_n}{b_n}\} \)is converges since one of them is always negative.

Therefore, if\(\{ {a_n}\} \)and \(\{ {b_n}\} \)are divergent, then\(\{ {a_n}{b_n}\} \)is convergent.