determine if the series converges or diverges. use any method, and give a reason for your answer. ∑n = 1 to…

determine if the series converges or diverges. use any method, and give a reason for your answer. ∑n = 1 to ∞ 6 / (n * n - th root of n) select the correct choice below and, if necessary, fill in the answer box to complete your choice. since limn→∞ an / bn = 1, where an = 6 / (n * n - th root of n) and bn = 6 / n, both series have positive terms, and the series ∑n = 1 to ∞ 6 / n diverges, the given series diverges by the limit comparison test. a. ∑n = 1 to ∞ 6 / n diverges, the given series diverges by the limit comparison test. b. the series diverges because the limit used in the nth - term test is. c. since limn→∞ an / bn = 1, where an = 6 / (n * n - th root of n) and bn = 6 / n, both series have positive terms, and the series ∑n = 1 to ∞ 6 / n converges, the given series converges by the limit comparison test.

determine if the series converges or diverges. use any method, and give a reason for your answer. ∑n = 1 to ∞ 6 / (n * n - th root of n) select the correct choice below and, if necessary, fill in the answer box to complete your choice. since limn→∞ an / bn = 1, where an = 6 / (n * n - th root of n) and bn = 6 / n, both series have positive terms, and the series ∑n = 1 to ∞ 6 / n diverges, the given series diverges by the limit comparison test. a. ∑n = 1 to ∞ 6 / n diverges, the given series diverges by the limit comparison test. b. the series diverges because the limit used in the nth - term test is. c. since limn→∞ an / bn = 1, where an = 6 / (n * n - th root of n) and bn = 6 / n, both series have positive terms, and the series ∑n = 1 to ∞ 6 / n converges, the given series converges by the limit comparison test.

Answer

Explanation:

Step1: Recall p - series test

The series $\sum_{n = 1}^{\infty}\frac{1}{n}$ is a harmonic series (a p - series with $p = 1$), and it diverges.

Step2: Apply Limit - Comparison Test

We are given $a_n=\frac{6}{n\sqrt[n]{n}}$ and $b_n=\frac{6}{n}$. Calculate $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{\frac{6}{n\sqrt[n]{n}}}{\frac{6}{n}}=\lim_{n\rightarrow\infty}\frac{1}{\sqrt[n]{n}}$. We know that $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$. So $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=1$. Since $\sum_{n = 1}^{\infty}b_n=\sum_{n = 1}^{\infty}\frac{6}{n}=6\sum_{n = 1}^{\infty}\frac{1}{n}$ diverges and $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=1$ with $a_n>0$ and $b_n>0$ for all $n\geq1$, by the Limit - Comparison Test, the series $\sum_{n = 1}^{\infty}\frac{6}{n\sqrt[n]{n}}$ diverges.

Answer:

A. Since $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=1$, where $a_n = \frac{6}{n\sqrt[n]{n}}$ and $b_n=\frac{6}{n}$, both series have positive terms, and the series $\sum_{n = 1}^{\infty}\frac{6}{n}$ diverges, the given series diverges by the Limit Comparison Test.