Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In problems 1-6, determine the convergence set of the given power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{3}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}{\mathit{x}}^{\mathbf{n}}$

Step-by-Step Solution

Answer

The set is, $x\in (-\infty ,\infty )$.

 

Step 1:To Find the Radius of convergence

Using the ratio test to determine the radius of convergence.

 $\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{3^n}{n!}}{\frac{3^{n+1}}{n+1!}}\right|\\ =\underset{n\to \infty }{\lim }|\frac{3^n}{3^{n+1}}\times \frac{(n+1)!}{n!}|\\ =\underset{n\to \infty }{\lim }|\frac{(n+1)n!}{n!}\times \frac{3^n}{3\times 3^n}|\\ =\underset{n\to \infty }{\lim }\left|\frac{n+1}{3}\right|\\ =\infty \end{array}$

The radius of convergence is $\infty$, therefore the series is convergent over the complete real line.

$\left|x\right|<\infty$

Step 2:Find the set of convergence

The convergent set for the given power series is

$x\in (-\infty ,\infty )$