Question

Use definition 2 directly to prove that \(\mathop {\lim }\limits_{n \to \infty } {r^n} = 0\)when\(\left| r \right| < 1\)

Short answer

Using the 2^nd definition of limit of a sequence, it is proved that\(\mathop {\lim }\limits_{n \to \infty } {r^n} = 0\)when \(\left| r \right| < 1\).

Step-by-step solution

Writing the 2nd definition of limit of a sequence:

A sequence {\({a_n}\)} has a limit \(L\)and we write \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\)or \({a_n} \to L\)as \(n \to \infty \)

If for every \(\varepsilon > 0\)there is a corresponding integer\(N\)such that if \(n > N\)then \(\left| {{a_n} - L} \right| < \varepsilon \)

Substituting \({r^n}\)in place of \({a_n}\):-

Consider\(\varepsilon > 0\), then we can write

\(\left| {{a_n} - L} \right| = \left| {{r^n} - 0} \right|\)

Since, there is no limit, \(L = 0.\)

\(\begin{aligned}{} \Rightarrow \left| {{a_n} - L} \right| = \left| {{r^n}} \right|\\ \Rightarrow \left| {{r^n}} \right| = {1^n}\end{aligned}\)

Since, we have \(\left| r \right| < 1\)from the question.

\( \Rightarrow \left| {{a_n} - L} \right| < 1,\forall n = 1,2,3,.....\)

Considering ‘\(N\)’ as an integer when \(\varepsilon  = 1\)(i.e., \(\varepsilon  > 0\)) and \(n \ge 1 = N\):-

For\(\varepsilon = 1( > 0)\)

\( \Rightarrow \left| {{a_n} - L} \right| < 1{\rm{ }}\)becomes,

\( \Rightarrow \left| {{a_n} - L} \right| < \varepsilon ,\forall n \ge N\)

Thus, By the definition,

\(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\)

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } {r^n} = 0\)where\(\left| r \right| < 1.\)

Therefore,

It is proved that \(\mathop {\lim }\limits_{n \to \infty } {r^n} = 0\)when \(\left| r \right| < 1\).