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Essential Calculus: Early Transcendentals
Found in: Page 434
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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

Step by Step Solution


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\(n \to \infty \).

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