Question

determine if the series converges or diverges. give a reason for your answer
sum _ { n = 1 } ^ { infty } \frac { 1 } { 6 sqrt { n } + sqrt 3 { n } }
choose the correct choice below
a. the limit comparison test with ( sum _ { n = 1 } ^ { infty } \frac { 1 } { sqrt { n } } ) shows that the series diverges
b. the limit comparison test with ( sum _ { n = 1 } ^ { infty } \frac { 1 } { sqrt 3 { n } } ) shows that the series converges
c. the limit comparison test with ( sum _ { n = 1 } ^ { infty } \frac { 1 } { sqrt 3 { n } } ) shows that the series diverges
d. the limit comparison test with ( sum _ { n = 1 } ^ { infty } \frac { 1 } { sqrt { n } } ) shows that the series converges

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 \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}} = L\), 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

ightarrow\infty}\frac{a_{n}}{b_{n}}\)

$$
LATEXBLOCK0
$$

As \(n
ightarrow\infty\), \(n^{-\frac{1}{6}}=\frac{1}{n^{\frac{1}{6}}}
ightarrow0\). So \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{1}{6}\) (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 \(\lim_{n
ightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{1}{6}\in(0,\infty)\) 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