Question

Use the ratio test to find whether the following series converge or diverge:

21.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{5}}^{\mathbf{n}}{\left(n!\right)}^{\mathbf{2}}}{\left(2n\right)\mathbf{!}}$

Short answer

The series $\sum _{n=0}^{\infty }\frac{5^n{\left(n!\right)}^2}{\left(2n\right)!}$is divergent.

Step-by-step solution

Process of ratio test

Apply ratio test in the given series by using${\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$androle="math" localid="1664187569161" $\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{\rho }}_{\mathbf{n}}$, where${\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$ is the${\left(n+1\right)}^{\mathbf{t}\mathbf{h}}$ term of the series and${\mathit{a}}_{\mathbf{n}}$ is the${\mathit{n}}^{\mathbf{t}\mathbf{h}}$ term. If $\mathit{\rho }\mathbf{<}\mathbf{1}$, then the series converges. If role="math" $\mathit{\rho }\mathbf{>}\mathbf{1}$, then the series diverges.

Apply the ratio test

The given series is $\sum _{n=0}^{\infty }\frac{5^n{\left(n!\right)}^2}{\left(2n\right)!}$.

So, $a_{n+1}=\frac{5^{n+1}{\left[\left(n+1\right)!\right]}^2}{\left[2\left(n+1\right)\right]!}$, and $a_n=\frac{5^n{\left(n!\right)}^2}{\left(2n\right)!}$.

Obtain the value of ${\rho }_n=\left|\frac{a_{n+1}}{a_n}\right|$.

$\begin{array}{rcl}{\rho }_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{5^{n+1}{\left[\left(n+1\right)!\right]}^2}{\left[2\left(n+1\right)\right]!}÷\frac{5^n{\left(n!\right)}^2}{\left(2n\right)!}\right|\\ & =& \frac{5^{n+1}{\left[\left(n+1\right)!\right]}^2}{\left[2\left(n+1\right)\right]!}\times \frac{\left(2n\right)!}{5^n{\left(n!\right)}^2}\\ & =& \frac{5{\left(n+1\right)}^2}{\left(2n+2\right)\left(2n+1\right)}\\ & =& \frac{5\left(n+1\right)}{2\left(2n+1\right)}\end{array}$

Solve the limit

Now,$\rho =\underset{n\to \infty }{\lim }{\rho }_n$ is calculated as follows:

role="math" $\begin{array}{rcl}\rho & =& \underset{n\to \infty }{\lim }\frac{5\left(n+1\right)}{2\left(2n+1\right)}\\ & =& \underset{n\to \infty }{\lim }\frac{5n\left(1+{\displaystyle \frac{1}{n}}\right)}{2n\left(2+{\displaystyle \frac{1}{n}}\right)}\\ & =& \underset{n\to \infty }{\lim }\frac{5\left(1+{\displaystyle \frac{1}{n}}\right)}{2\left(2+{\displaystyle \frac{1}{n}}\right)}\\ & =& \frac{5\left(1+0\right)}{2\left(2+0\right)}\\ \rho & =& \frac{5}{4}\\ & & \end{array}$

Here, $\rho >1$, therefore the series diverges.