Question

Use the preliminary test to decide whether the following series are divergent or require further testing. Careful:Do notsay that a series is convergent; the preliminary test cannot decide this.

6.$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{!}}$

Short answer

The series requires further testing.

Step-by-step solution

Use the preliminary test

Consider the series:

$\sum _{\mathrm{n}=1}^{\infty }\frac{\mathrm{n}!}{(\mathrm{n}+1)!}$

Use the preliminary test to see if this series diverges or converges, where ${\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}!}{(\mathrm{n}+1)!}$.

So, $\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{a}}_{\mathrm{n}}=\underset{\mathrm{n}\to \infty }{\lim }\frac{\mathrm{n}!}{(\mathrm{n}+1)!}$.

Check the parameter

It is known that $(\mathrm{n}+1)!=(\mathrm{n}+1).\mathrm{n}!$. Hence,

$\begin{array}{rcl}\underset{n\to \infty }{\lim }{a}_n& =& \underset{n\to \infty }{\lim }\frac{n!}{(n+1)!}\\ & =& \underset{n\to \infty }{\lim }\frac{1}{(n+1)}\\ & =& \frac{1}{\infty }\\ & =& 0\end{array}$

Because $\underset{\mathrm{n}\to \infty }{\lim }{\mathrm{a}}_{\mathrm{n}}=0$, this means that preliminary testing is not enough and the series will need to be further tested using other methods.