Question

Prove that if \(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)and {\({b_n}\)} is bounded, then \(\mathop {\lim }\limits_{n \to \infty } ({a_n}{b_n}) = 0.\)

Short answer

Using the definition of aboundedsequence, it is obtained that\(\mathop {\lim }\limits_{n \to \infty } ({a_n}{b_n}) = 0\) when\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)and {\({b_n}\)} is bounded.

Step-by-step solution

Given data:-

\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)and{\({b_n}\)}is bounded. Since {\({b_n}\)} is bounded, From the definition,we know that there is a positive number \(M\)such that,

\(\left| {{b_n}} \right| \le M\)for all \(n \ge 1.\)

We can write this as,

\(\left| {{a_n}} \right| \cdot \left| {{b_n}} \right| \le \left| {{a_n}} \right| \cdot M\), \(\forall n \ge 1.\)

\(L\)is the limit of the sequence {\({a_n}\)},\(f\)is the function continuous at \(L.\)

ApplyingDefinition 2, limit of a sequence to the given limit :-

\(\begin{aligned} &\mathop {\lim }\limits_{n \to \infty } ({a_n}{b_n})\\ &\Rightarrow \left| {{a_n}{b_n} - 0} \right| = \left| {{a_n}{b_n}} \right|\\ &\Rightarrow \left| {{a_n}} \right| \cdot \left| {{b_n}} \right| \le \left| {{a_n}} \right| \cdot M\\ &\Rightarrow \left| {{a_n} - 0} \right| \cdot M \le \varepsilon '\end{aligned}\)

Since, if \(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\) for every \(\varepsilon > 0\), there is a corresponding integer \(N\)such that,

\(\left| {{a_n} - 0} \right| < \varepsilon '\),for all\(n < N\).

Then,

\(\left| {{a_n} - 0} \right| \cdot M < \varepsilon '\)where\(\varepsilon ' = \varepsilon \cdot M > 0.\)

Therefore, \(\left| {{a_n}{b_n} - 0} \right| < \varepsilon '\),so the sequence {\({a_n}{b_n}\)} has limit zero as \(n \to \infty \).

Hence it is proved that,\(\mathop {\lim }\limits_{n \to \infty } ({a_n}{b_n}) = 0\).