Chapter 10.5, Problem 46E

a.

To determine

To find: the interval of convergence of the series.

Answer to Problem 46E

Interval of convergence is $\left\{4\right\}$ .

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=0}^{\infty }n!{(x-4)}^n}$

Formula used:

Power series about $x=a$ , is given by ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$

If $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|<1$ , then the series converges. And if $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|>1$ , then the series diverges.

For convergent series the limit will result in $\frac{|x-a|}{R}<1$ , where R is the radius of convergence and $|x-a|<R$ will give the interval of convergence.

Calculation:

Applying limit we get,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{(n+1)!{(x-4)}^{n+1}}{n!{(x-4)}^n}\right|\\ \underset{n\to \infty }{\lim }\left|(n+1)(x-4)\right|\\ =\infty >1\end{array}$

As $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|>1$ , thus the series only converges for $x=4$ and diverges at all points.

The radius of convergence, $R=0$ .

Therefore the interval of convergence is {4}.

b.

To determine

To find: Values of x for which the series converges absolutely.

Answer to Problem 46E

The series converges absolutely for $x=4$ .

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=0}^{\infty }n!{(x-4)}^n}$

Formula used:

Power series about $x=a$ , is given by ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$

For convergent series the limit will result in $\frac{|x-a|}{R}<1$ , where R is the radius of convergence and $|x-a|<R$ will give the interval of convergence.

The series converges absolutely for the values of x in the interval of convergence.

Calculation:

Applying limit we get,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{(n+1)!{(x-4)}^{n+1}}{n!{(x-4)}^n}\right|\\ \underset{n\to \infty }{\lim }\left|(n+1)(x-4)\right|\\ =\infty >1\end{array}$

As $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|>1$ , thus the series only converges for $x=4$ and diverges at all points.

The radius of convergence, $R=0$

Therefore the interval of convergence is {4}.

The series converges absolutely for $x=4$ .

c.

To determine

To find: Values of x for which the series converges conditionally.

Answer to Problem 46E

The series does not converge conditionally at any point.

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=0}^{\infty }n!{(x-4)}^n}$

Formula used:

Power series about $x=a$ , is given by ${\displaystyle \sum _{n=0}^{\infty }{c}_n{(x-a)}^n}$

For convergent series the limit will result in $\frac{|x-a|}{R}<1$ , where R is the radius of convergence and $|x-a|<R$ will give the interval of convergence.

The series converges at end points then the series converges conditionally at that points.

Calculation:

Applying limit we get,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{(n+1)!{(x-4)}^{n+1}}{n!{(x-4)}^n}\right|\\ \underset{n\to \infty }{\lim }\left|(n+1)(x-4)\right|\\ =\infty >1\end{array}$

As $\underset{n\to \infty }{\lim }\left|\frac{u_{n+1}}{u_n}\right|>1$ , thus the series only converges for $x=4$ and diverges at all points.

The radius of convergence, $R=0$

Therefore the interval of convergence is {4}.

The series does not converge conditionally at any point.