use the root test to determine if the following series converges or diverges\n\n\\( \\sum _ { n = 15 } ^ {…

use the root test to determine if the following series converges or diverges\n\n\\( \\sum _ { n = 15 } ^ { \\infty } \\frac { ( n \\cdot ( 1 + n ) ) ^ { 2 n } } { 6 ^ { - 9 n } } \\)\n\nsince the limit resulting from the root test is \\( \\frac { 1 } { 36 } \\), which is \\( < 1 \\), the root test shows the series converges\n(simplify your answer. type an exact answer.)

use the root test to determine if the following series converges or diverges\n\n\\( \\sum _ { n = 15 } ^ { \\infty } \\frac { ( n \\cdot ( 1 + n ) ) ^ { 2 n } } { 6 ^ { - 9 n } } \\)\n\nsince the limit resulting from the root test is \\( \\frac { 1 } { 36 } \\), which is \\( < 1 \\), the root test shows the series converges\n(simplify your answer. type an exact answer.)

Answer

Explanation:

Step1: Apply the root test formula

For a series (\sum_{n = 1}^{\infty}a_{n}), the root - test is (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}). Here (a_{n}=\frac{(n + 1)n}{36^{n}}). Then (\sqrt[n]{\vert a_{n}\vert}=\sqrt[n]{\frac{(n + 1)n}{36^{n}}}). Using the property (\sqrt[n]{\frac{M}{N}}=\frac{\sqrt[n]{M}}{\sqrt[n]{N}}), we have (\sqrt[n]{\vert a_{n}\vert}=\frac{\sqrt[n]{n(n + 1)}}{36}).

Step2: Evaluate the limit

We know that (\lim_{n\rightarrow\infty}\sqrt[n]{n}=1) and (\lim_{n\rightarrow\infty}\sqrt[n]{n + 1}=1) (by the formula (\lim_{n\rightarrow\infty}n^{\frac{1}{n}}=e^{\lim_{n\rightarrow\infty}\frac{\ln n}{n}}), and using L'Hopital's rule (\lim_{n\rightarrow\infty}\frac{\ln n}{n}=\lim_{n\rightarrow\infty}\frac{1/n}{1}=0)). So (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=\frac{\lim_{n\rightarrow\infty}\sqrt[n]{n(n + 1)}}{36}=\frac{\lim_{n\rightarrow\infty}\sqrt[n]{n}\cdot\lim_{n\rightarrow\infty}\sqrt[n]{n + 1}}{36}). Substituting (\lim_{n\rightarrow\infty}\sqrt[n]{n}=1) and (\lim_{n\rightarrow\infty}\sqrt[n]{n + 1}=1) into the above formula, we get (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=\frac{1\times1}{36}=\frac{1}{36}).

Since (\frac{1}{36}<1), by the root - test, the series (\sum_{n = 1}^{\infty}\frac{(n + 1)n}{36^{n}}) converges.

Answer:

The series (\sum_{n = 1}^{\infty}\frac{(n + 1)n}{36^{n}}) converges.