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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.$${a_n} = \ln (n + 1) - \ln n$$

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:

\begin{aligned}{a_n} &= \ln (n + 1) - \ln n\\ &= \ln \left( {\frac{{n + 1}}{n}} \right)\\ &= \ln \left( {1 + \frac{1}{n}} \right)\end{aligned}

So, the limit:

\begin{aligned}\mathop {\lim }\limits_{n \to \infty } \left( {\ln \left( {1 + \frac{1}{n}} \right)} \right) &= \ln 1\\ &= 0\end{aligned}

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