Chapter 10, Problem 13RE

(a)

To determine

The radius of the convergence of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$

Answer to Problem 13RE

The radius or convergence for it is 0. $R=0$

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$

Formula used:

  $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{n!\left|x^{2n}\right|}{2^n}.\frac{2^{n-1}}{\left(n-1\right)!.\left|x^{2n-2}\right|}$

Calculation:

Check for absolute convergence using ratio test. $\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{n!\left|x^{2n}\right|}{2^n}.\frac{2^{n-1}}{\left(n-1\right)!.\left|x^{2n-2}\right|}$

  $\begin{array}{l}=\underset{n\to \infty }{\lim }n\frac{\left|x^2\right|}{2}\\ =\infty ,\text{if }x\ne 0\\ =0,\text{if }x=0\end{array}$

Therefore, the series is convergent only for x = 0. Therefore, the radius or convergence for it is 0. $R=0$

Conclusion:

The radius or convergence for it is 0. $R=0$

(b)

To determine

The interval of convergence of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$

Answer to Problem 13RE

The series is convergent only at $x=0$

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$

Formula used:

Ratio test is used.

Calculation:

Since,

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

Only for x = 0. Therefore, the series is convergent only at $x=0$

Conclusion:

The series is convergent only at $x=0$

(c)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$ converges absolutely.

Answer to Problem 13RE

The series is absolutely convergent only at $x=0$

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$

Formula used:

Ratio test is used.

Calculation:

Since,

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

Only for x = 0. Therefore, the series is absolutely convergent only at $x=0$

Conclusion:

The series is absolutely convergent only at $x=0$

(d)

To determine

The series ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$ converges conditionally.

Answer to Problem 13RE

There is no point of conditional convergence.

Explanation of Solution

Given information:

The given series is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{n!}{2^n}}{x}^{2n}$

Formula used:

Ratio test is used.

Calculation:

Since, the series is convergent only at x = 0, there is no point of conditional convergence.

Conclusion:

There is no point of conditional convergence.