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Q32E

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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.$${a_n} = \frac{{{{( - 3)}^n}}}{{n!}}$$

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$${a_n} = \frac{{{{( - 3)}^n}}}{{n!}}$$.

It can be written as$$\frac{{{{( - 3)}^n}}}{{n!}} = \frac{{{{( - 1)}^n}{{(3)}^n}}}{{n!}}$$

We know that$$n! > {a^n}$$so

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

Thus given sequence is convergent and converges to 0.