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Q17E

Expert-verified
Found in: Page 434

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Determine whether the sequence converges or diverges. If it converges, find the limit.$${a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}$$

The sequence converges and the value of limit is 0.

See the step by step solution

Definition

A sequence$$\left\{ {{a_n}} \right\}$$has the limit $$L$$and we write $$\mathop {\lim }\limits_{n \to \infty } {a_n} = L\;\;or\;\;{a_n} \to L$$as$$n \to \infty$$if we can make the terms$${a_n}$$as close to$$L$$as we like by taking$$n$$sufficiently large. If $$\mathop {\lim }\limits_{n \to \infty } {a_n}$$exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).

Simplify

Given a sequence $${a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}$$.

Evaluate$$\mathop {\lim }\limits_{n \to \infty } |{a_n}|$$ as:

\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{( - 1)}^n}}}{{2\sqrt n }}} \right| &= \frac{1}{2}\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{\sqrt n }}} \right)\\ &= \frac{1}{2} \cdot 0\\ &= 0\end{aligned}

Compute$$\mathop {\lim }\limits_{n \to \infty } {a_n}$$

Now using the theorem:

If$$\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| = 0$$, then$$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$$

\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {a_n} &= \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{( - 1)}^n}}}{{2\sqrt n }}} \right)\\ &= 0\end{aligned}.

So, the sequence $${a_n} = \frac{{{{( - 1)}^n}}}{{2\sqrt n }}$$ converges and the value of limit is 0.

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