Question

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

22.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{10}}^{\mathbf{n}}}{{\left(n!\right)}^{\mathbf{2}}}$

Short answer

The series$\sum _{n=1}^{\infty }\frac{{10}^n}{{\left(n!\right)}^2}$ is convergent.

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$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}{\mathit{\rho }}_{\mathbf{n}}$, 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 role="math" $\mathit{\rho }\mathbf{<}\mathbf{1}$, then the series converges. If role="math" $\mathit{\rho }\mathbf{>}\mathbf{1}$, then the series diverges.

Apply the ratio test

The given series is$\sum _{n=1}^{\infty }\frac{{10}^n}{{\left(n!\right)}^2}$.

So, $a_{n+1}=\frac{{10}^{n+1}}{{\left[\left(n+1\right)!\right]}^2}$, and localid="1664185894811" $a_n=\frac{{10}^n}{{\left(n!\right)}^2}$.

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

$\begin{array}{rcl}{\rho }_n& =& \left|\frac{a_{n+1}}{a_n}\right|\\ & =& \left|\frac{{10}^{n+1}}{{\left[\left(n+1\right)!\right]}^2}÷\frac{{10}^n}{{\left(n!\right)}^2}\right|\\ & =& \frac{{10}^{n+1}}{{\left[\left(n+1\right)!\right]}^2}\times \frac{{\left(n!\right)}^2}{{10}^n}\\ & =& \frac{10}{{\left(n+1\right)}^2}\end{array}$

Solve the limit

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

role="math" localid="1664186203147" $\begin{array}{rcl}\rho & =& \underset{n\to \infty }{\lim }\frac{10}{{\left(n+1\right)}^2}\\ & =& \frac{10}{{\left(\infty +1\right)}^2}\\ & =& \frac{10}{\infty }\\ & =& 0\end{array}$

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