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Test the series for convergence or divergence. \

n=1(3n1+8n)n\sum_{n=1}^{\infty}\left(\dfrac{3 n}{1+8 n}\right)^n


Answered 2 years ago
Answered 2 years ago
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We look at the series:

n=1(3n1+8n)n\sum_{n=1}^{\infty} \bigg( \dfrac{3n}{1 + 8n}\bigg)^n.

We can write this as n=1an\sum_{n=1}^{\infty} a_n where an=(3n1+8n)na_n = ( \frac{3n}{1 + 8n})^n. Because we have a fairly simple expression raised to the nthn^{th} power we will use the Root Test. We want to look at limnann\lim_{n \rightarrow \infty} \sqrt[n]{\left| a_n \right|}. We have:

limnann=limn(3n1+8n)nn\lim_{n \rightarrow \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \rightarrow \infty} \sqrt[n]{ \bigg( \dfrac{3n}{1 + 8n}\bigg)^n}

=limn3n1+8n=\lim_{n \rightarrow \infty} \dfrac{3n}{1+8n}

=limn31n+8=\lim_{n \rightarrow \infty} \dfrac{3}{\frac{1}{n} + 8}

=30+8=\dfrac{3}{0 + 8}

=38<1=\dfrac{3}{8} < 1

The Root Test tells us that if limnann<1\lim_{n \rightarrow \infty} \sqrt[n]{\left| a_n \right|} < 1 then n=1an\sum_{n=1}^{\infty} a_n converges, so we conclude our series n=1an=n=1(3n1+8n)n\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} ( \frac{3n}{1 + 8n})^n converges.

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