Question

use the root test to determine if the following series converges or diverges

\\( \sum _ { n = 15 } ^ { \infty } \frac { ( n \cdot ( 1 + n ) ) ^ { 2 n } } { 6 ^ { - 9 n } } \\)

since the limit resulting from the root test is \\( \frac { 1 } { 36 } \\), which is \\( < 1 \\), the root test shows the series converges
(simplify your answer. type an exact answer.)

Explanation:

Step1: Apply the root test formula

For a series \(\sum_{n = 1}^{\infty}a_{n}\), the root - test is \(\lim_{n
ightarrow\infty}\sqrt[n]{\vert a_{n}\vert}\). 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 \(\lim_{n
ightarrow\infty}\sqrt[n]{n}=1\) and \(\lim_{n
ightarrow\infty}\sqrt[n]{n + 1}=1\) (by the formula \(\lim_{n
ightarrow\infty}n^{\frac{1}{n}}=e^{\lim_{n
ightarrow\infty}\frac{\ln n}{n}}\), and using L'Hopital's rule \(\lim_{n
ightarrow\infty}\frac{\ln n}{n}=\lim_{n
ightarrow\infty}\frac{1/n}{1}=0\)).
So \(\lim_{n
ightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=\frac{\lim_{n
ightarrow\infty}\sqrt[n]{n(n + 1)}}{36}=\frac{\lim_{n
ightarrow\infty}\sqrt[n]{n}\cdot\lim_{n
ightarrow\infty}\sqrt[n]{n + 1}}{36}\).
Substituting \(\lim_{n
ightarrow\infty}\sqrt[n]{n}=1\) and \(\lim_{n
ightarrow\infty}\sqrt[n]{n + 1}=1\) into the above formula, we get \(\lim_{n
ightarrow\infty}\sqrt[n]{\vert a_{n}\vert}=\frac{1\times1}{36}=\frac{1}{36}\).

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.