Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{!}}{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{)}}^{\mathbf{n}}$

Step-by-Step Solution

Answer

 The set is, $x\in (-3,-1)$

Step 1:To Find the Radius of convergence

Use the ratio test to determine the radius of convergence (with ):$a_n=\frac{(n+2)!}{n!}$

 $\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{(n-2)!}{n!}}{\frac{(n+3)!}{(n+1)!}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{(n+2)!}{(n+3)!}\cdot \frac{(n+1)!}{n!}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{(n+2)!}{(n+3)(n+2)!}\times \frac{(n+1)n!}{n!}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{(n+1)}{n+3}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{(1+(1/n\left)\right)}{1+(3/n)}\right|\\ =1\end{array}$

The radius of convergence is 1, therefore convergent set for the given power series is.$|x+2|<1$

Step 2: Find the set of convergence

To completely identify the convergence set, we have to check whether the boundary points -1 and -3 are included in the set or not.

 Checking at$x=-1$ , by substituting the value ofx  by -1 .

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{(n+2)!}{n!} {\left(x+2\right)}^n=\sum _{n=0}^{\infty }\frac{(n+2)!}{n!} {\left(-1+2\right)}^n\\ =\sum _{n=0}^{\infty }\frac{(n+2)!}{n!} {\left(1\right)}^n\\ =\sum _{n=0}^{\infty }\frac{(n+2)(n+1)n!}{n!}\\ =\sum _{n=0}^{\infty }(n+2)(n+1)\\ =\infty \end{array}$

The above series is divergent, thus the point  -1 is excluded from the convergent set

Similarly, checking at $x=-3$ by substituting the value of x by  -3.

 $\begin{array}{c}\sum _{n=0}^{\infty }\frac{(n+2)!}{n!} {\left(x+2\right)}^n=\sum _{n=0}^{\infty }\frac{(n+2)!}{n!} {\left(-3+2\right)}^n\\ =\sum _{n=0}^{\infty }\frac{(n+2)!}{n!}\\ =\sum _{n=0}^{\infty }\frac{(n+2)(n+1)n!}{n!}\\ =\sum _{n=0}^{\infty }{(-1)}^n(n+2)(n+1)\end{array}$

The above series is divergent, thus the point -3 is not included in the convergent set. The convergent set for the given power series is.$x\in (-3,-1)$