Question

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

23.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}\mathbf{!}}{{\mathbf{100}}^{\mathbf{n}}}$

Short answer

The series $\sum _{n=1}^{\infty }\frac{n!}{{100}^n}$divergence.

Step-by-step solution

Process of ratio test

Apply ratio test in the given series by using${\mathit{\rho }}_{\mathbf{n}}\mathbf{=}\left|\frac{a_{n+1}}{a_n}\right|$and$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{\rho }}_{\mathbf{n}}$, 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 $\mathit{\rho }\mathbf{>}\mathbf{1}$, then the series diverges.

Apply the ratio test

The given series is $\sum _{n=1}^{\infty }\frac{n!}{{100}^n}$.

So, $a_{n+1}=\frac{\left(n+1\right)!}{{100}^{n+1}}$, role="math" $a_n=\frac{n!}{{100}^n}$.

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

$\begin{array}{rcl}{\rho }_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{\left(n+1\right)!}{{100}^{n+1}}÷\frac{n!}{{100}^n}\right|\\ & =& \frac{\left(n+1\right)!}{{100}^{n+1}}\times \frac{{100}^n}{n!}\\ & =& \frac{\left(n+1\right)}{100}\end{array}$

Solve the limit

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

$\begin{array}{rcl}\rho & =& \underset{n\to \infty }{\lim }\frac{n+1}{100}\\ & =& \frac{\infty }{100}\\ & =& \infty \end{array}$

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