determine if the series converges or diverges. give a reason for your answer\n sum _ { n = 1 } ^ { infty }…

Explanation:

Step1: Recall the limit - comparison test formula

Let (\sum_{n = 1}^{\infty}a_{n}) and (\sum_{n=1}^{\infty}b_{n}) be series with (a_{n}\gt0) and (b_{n}\gt0) for all (n). Then (\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}} = L), where (0\lt L\lt\infty). The series (\sum_{n = 1}^{\infty}a_{n}) and (\sum_{n=1}^{\infty}b_{n}) either both converge or both diverge. Let (a_{n}=\frac{1}{6\sqrt{n}+\sqrt[3]{n}}) and (b_{n}=\frac{1}{\sqrt{n}}).

Step2: Calculate the limit (\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}})

[ \begin{align*} \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}&=\lim_{n\rightarrow\infty}\frac{\frac{1}{6\sqrt{n}+\sqrt[3]{n}}}{\frac{1}{\sqrt{n}}}\ &=\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{6\sqrt{n}+\sqrt[3]{n}}\ &=\lim_{n\rightarrow\infty}\frac{1}{6 + n^{-\frac{1}{6}}}\ \end{align*} ] As (n\rightarrow\infty), (n^{-\frac{1}{6}}=\frac{1}{n^{\frac{1}{6}}}\rightarrow0). So (\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{1}{6}) (a positive finite number).

Step3: Recall the (p -)series test

The (p -)series (\sum_{n = 1}^{\infty}\frac{1}{n^{p}}) converges if (p>1) and diverges if (p\leq1). For the series (\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}=\sum_{n=1}^{\infty}\frac{1}{n^{\frac{1}{2}}}), where (p = \frac{1}{2}\leq1), so (\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}) diverges.

Since (\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{1}{6}\in(0,\infty)) and (\sum_{n = 1}^{\infty}b_{n}=\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}) diverges, by the limit - comparison test, (\sum_{n = 1}^{\infty}a_{n}=\sum_{n = 1}^{\infty}\frac{1}{6\sqrt{n}+\sqrt[3]{n}}) diverges.

Answer:

A. The limit comparison test with (\sum_{n = 1}^{\infty}\frac{1}{\sqrt{n}}) shows that the series diverges