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# Test the series for convergence or divergence. \$\sum_{n=1}^{\infty}\left(\dfrac{3 n}{1+8 n}\right)^n$

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We look at the series:

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

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

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

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

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

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

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

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

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