Use the limit comparison test to determine if ∑n = 1∞6 n45+3 n+5 n14−4

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Use the limit comparison test to determine if ∑n = 1∞6 n45+3 n+5 n14−4 converges or diverges, and justify your answer. Answer Attempt 1 out of 3 Apply the comparison test with the series ∑n = 1∞1 np where p = . If an = 6 n45+3 n+5 n14−4 and bn = 1 np, then limn→∞an bn = . Since an, bn > 0 and the limit is a finite and positive (non-zero) number, the limit comparison test applies. ∑n = 1∞1 np since a p-series will if and only if . Therefore, ∑n = 1∞6 n45+3 n+5 n14−4

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Use the limit comparison test to determine if

\sum_{n = 1}^{\infty} \frac{6 n^{\frac{4}{5}} + 3}{n + 5 n^{\frac{1}{4}} - 4}

converges or diverges, and justify your answer. Answer Attempt 1 out of 3

Apply the comparison test with the series

\sum_{n = 1}^{\infty} \frac{1}{n^{p}}

where

p = ◻

. If

a_{n} = \frac{6 n^{\frac{4}{5}} + 3}{n + 5 n^{\frac{1}{4}} - 4}

and

b_{n} = \frac{1}{n^{p}}

, then

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = ◻

. Since

a_{n}, b_{n} > 0

and the limit is a finite and positive (non-zero) number, the limit comparison test applies.

\sum_{n = 1}^{\infty} \frac{1}{n^{p}}

since a p-series will if and only if .Therefore,

\sum_{n = 1}^{\infty} \frac{6 n^{\frac{4}{5}} + 3}{n + 5 n^{\frac{1}{4}} - 4}

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