Question

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

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

Short answer

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

Step-by-step solution

 Step 1: Process of ratio test

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

Apply the ratio test

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

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

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

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

Solve the limit

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

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

Here, $\rho <1$, therefore the series converges.