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Essential Calculus: Early Transcendentals
Found in: Page 453
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 series is convergent or divergent:\(\sum\limits_{n = 1}^\infty {\frac{{n + 5}}{{\sqrt(3){{{n^7} + {n^2}}}}}} \).

The series \(\sum\limits_{n = 1}^\infty {\frac{{n + 5}}{{\sqrt(3){{{n^7} + {n^2}}}}}} \) is convergent.

See the step by step solution

Step by Step Solution

From the series letting \({a_n}\) and \({b_n}\):

We have the series \(\sum\limits_{n = 1}^\infty {\frac{{n + 5}}{{\sqrt(3){{{n^7} + {n^2}}}}}} \)

Take \({a_n} = \frac{{n + 5}}{{\sqrt(3){{{n^7} + {n^2}}}}}\), \({b_n} = \frac{n}{{\sqrt(3){{{n^7}}}}}\)

\( \Rightarrow {b_n} = \frac{1}{{{n^{\frac{4}{3}}}}}\)

Using Limit comparison Test:

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n + 5}}{{\sqrt(3){{{n^7} + {n^2}}}}}{n^{\frac{4}{3}}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^{\frac{7}{3}}} + 5{n^{\frac{4}{3}}}}}{{{{({n^7} + {n^2})}^{\frac{1}{3}}}}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{1 + 5{n^{\frac{4}{3} - \frac{7}{3}}}}}{{{{(1 + \frac{1}{{{n^5}}})}^{\frac{1}{3}}}}}\)

\(\mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{5}{n}}}{{{{(1 + \frac{1}{{{n^5}}})}^{\frac{1}{3}}}}}\)

Comparison Tests:

Now, \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \frac{{1 + \frac{5}{n}}}{{{{(1 + \frac{1}{{{n^5}}})}^{\frac{1}{3}}}}} = 1 > 0\)

Since, limit exists and \(\sum\limits_{n = 1}^\infty {\frac{1}{{{n^{\frac{4}{3}}}}}} \) converges by p-series test with \(p = \frac{4}{3} > 1\).

Therefore, the given series also converges by limit comparison test.

Hence, the series is convergent.

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