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Q30E

Expert-verified
Found in: Page 453

### Essential Calculus: Early Transcendentals

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

# 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

## 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.