Question

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

25.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{e}}^{\mathbf{n}}}{\sqrt{\mathbf{n}\mathbf{!}}}$

Short answer

The series $\sum _{n=0}^{\infty }\frac{e^n}{\sqrt{n!}}$ 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$\mathit{\rho }\mathbf{=}\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{\rho }}_{\mathbf{n}}$,where${\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}$ is thedata-custom-editor="chemistry" ${\left(\mathit{n}+1\right)}^{\mathbf{t}\mathbf{h}}$ 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 $\sum _{n=0}^{\infty }\frac{e^n}{\sqrt{n!}}$.

So, role="math" $a_{n+1}=\frac{e^{n+1}}{\sqrt{\left(n+1\right)!}}$, and role="math" localid="1664177251873" $a_n=\frac{e^n}{\sqrt{n!}}$.

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

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

Solve the limit

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

$\begin{array}{rcl}\rho & =& \underset{n\to \infty }{\lim }\frac{e}{\sqrt{n+1}}\\ & =& \frac{e}{\sqrt{\infty +1}}\\ & =& \frac{e}{\infty }\\ & =& 0\end{array}$

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