Chapter 10, Problem 48RE

To determine

To check: whether the Series ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n{3}^n}{n^n}}$ converges absolutely, converges conditionally, or diverges.

Answer to Problem 48RE

The given series, ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n{3}^n}{n^n}}$ converges absolutely.

Explanation of Solution

Given:

The series is, ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n{3}^n}{n^n}}$ .

Calculation:

The given series is, ${\displaystyle \sum _{n=1}^{\infty }\frac{2^n{3}^n}{n^n}}$

Taking ratio test for absolute convergence,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{2^n{3}^n}{n^n}\cdot \frac{{\left(n-1\right)}^{n-1}}{2^{n-1}\cdot 3^{n-1}}\\ =\underset{n\to \infty }{\lim }\frac{6{\left(n-1\right)}^{n-1}}{n^n}=0\end{array}$

As, $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0$ ,

Thus the series converges absolutely for all real x. Therefore, the series converges absolutely for interval $\left(-\infty ,\infty \right).$