Chapter 10, Problem 4RE

(a)

To determine

the radius of the convergence ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Answer to Problem 4RE

The radius of convergence for is $\infty$ . $R=\infty$

Explanation of Solution

Given information:

The given function is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Formula used:

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

Calculation:

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

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{{\left|x-1\right|}^2}{\left(2n+1\right)\left(2n\right)}\\ ={\left|x-1\right|}^2\underset{n\to \infty }{\lim }\frac{1}{\left(2n+1\right)\left(2n\right)}\\ ={\left|x-1\right|}^2.0\\ =0\end{array}$

Therefore, the series is convergence for all real x. Therefore, the radius of convergence for is $\infty$ . $R=\infty$

Conclusion:

The radius of convergence for is $\infty$ . $R=\infty$

(b)

To determine

The radius of the convergence ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Answer to Problem 4RE

The series always converges for interval $\left(-\infty ,\infty \right)$

Explanation of Solution

Given information:

The given function is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Formula used:

Ratio test is used.

Calculation:

Since,

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

Therefore the series always absolutely for all real x

Since,

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

Therefore, the series always converges for interval $\left(-\infty ,\infty \right)$

Conclusion:

The series always converges for interval $\left(-\infty ,\infty \right)$

(c)

To determine

the series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Converges absolutely.

Answer to Problem 4RE

The series is always converges absolutely for interval $\left(-\infty ,\infty \right)$

Explanation of Solution

Given information:

The given function is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Formula used:

Ratio test is used.

Calculation:

Since,

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

Therefore the series always converges absolutely for all real x

Since,

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

Therefore, the series is always converges absolutely for interval $\left(-\infty ,\infty \right)$ .

Conclusion:

The series is always converges absolutely for interval $\left(-\infty ,\infty \right)$

(d)

To determine

the series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$ Converges conditionally.

Answer to Problem 4RE

There is no point of conditional convergence.

Explanation of Solution

Given information:

The given function is,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x-1\right)}^{2n-2}}{\left(2n-1\right)!}}$

Formula used:

Ratio test is used.

Calculation:

Since, the series always converges absolutely for all real x, there is no point of conditional convergence.

Conclusion:

There is no point of conditional convergence.