Question

Find the interval of convergence, including end-point tests:$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{n}\mathbf{!}\mathbf{)}}^{\mathbf{2}}{\mathbf{x}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}}$

Short answer

The interval of convergence is $-4<x<4$.

Step-by-step solution

Concept used to find the interval of convergence

The ratio test for convergence is expressed as follows:

$\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}\mathbf{<}\mathbf{1}$

The ratio test for divergence is expressed as follows:

$\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}\mathbf{>}\mathbf{1}$

Calculation to find the interval of the convergence of the series ∑n=1∞(n!)2xn(2n)!

term of the series is,$a_n=\frac{{(n!)}^2{x}^n}{\left(2n\right)!}$

The magnitude of $n^{\text{th}}$terms is:

$\left|a_n\right|=\frac{{(n!)}^2{x}^n}{\left(2n\right)!}$ …… (1)

The magnitude of $(n+1{)}^{th}$terms is calculated as follows:

$\left|a_{n+1}\right|=\frac{{\left(\right(n+1)!)}^2{x}^{n+1}}{(2n+2)!}$ …… (2)

Take the ratio of equation (2) y equation (1) as follows:

$\begin{array}{rcl}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}& =& \frac{\frac{{\left(\right(n+1)!)}^2{x}^{n+1}}{(2n+2)!}}{\frac{(n!)x^n}{\left(2n\right)!}}\\ & =& \frac{{\left(\right(n+1)!)}^2{x}^{n+1}\left(2n\right)!}{(2n+2)!{(n!)}^2{x}^n}\\ & =& \frac{{(n+1)}^{2}x}{(2n+2)(2n+1)}\\ & =& \frac{{\left(1+\frac{1}{n}\right)}^{2}x}{\left(2+\frac{2}{n}\right)\left(2+\frac{1}{n}\right)}\\ & & \end{array}$

Apply the limit in the above ratio as follows:

$$\begin{gathered} li{m}_{n\to \infty }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=li{m}_{n\to \infty }\left(\frac{{\left(1+\frac{1}{n}\right)}^{2}x}{\left(2+\frac{2}{n}\right)\left(2+\frac{1}{n}\right)}\right) \\ li{m}_{n\to \infty }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\left(\frac{x}{4}\right) \end{gathered}$$

The series converges for $li{m}_{n\to \infty }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}<1$when$\left|\frac{x}{4}\right|<1$ and the series is divergent when $\left|\frac{x}{4}\right|>1$.

Therefore, the interval is $-1<\frac{x}{4}<1$.

It is simplified as,

$$\begin{gathered} -1<\frac{x}{4}<1 \\ -4<x<4 \end{gathered}$$

For, $x=4$the series becomes:

$\sum _{n=1}^{\infty }\frac{{(n!)}^2{4}^{n+1}}{\left(2n\right)!}$

This series is divergent using the limit test.

For, $x=-4$the series becomes:

$\sum _{n=1}^{\infty }\frac{{(n!)}^2{(-4)}^{n+1}}{\left(2n\right)!}$

This is divergent series using the limit test.

Thus, the interval of converges is $-4<x<4$.