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Q26E

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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.$${a_n} = {2^{ - n}}\cos n\pi$$

Converges &$$\mathop {\lim }\limits_{n \to \infty } {a_n} = 0$$

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).

## Evaluate limit

Consider the sequence:

\begin{aligned}{a_n} &= {2^{ - n}}\cos n\pi \\ &= \frac{{\cos n\pi }}{{{2^n}}}\end{aligned}

Hence, we can find the limit:

\left\{{\begin{aligned}{\mathop {\lim }\limits_{n \to \infty }(\cos n\pi ) \in ( - 1,1)}\\{\mathop {\lim }\limits_{n \to \infty} \left( {{2^n}} \right) = \infty }\end{aligned}} \right.

And then$$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{\cos n\pi }}{{{2^n}}}} \right) = 0$$

So the sequence converges to zero as$$n \to \infty$$.