Chapter 10.4, Problem 44E

To determine

To find: the convergence or divergence of the series.

Answer to Problem 44E

The series is convergence

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=0}^{\infty }\frac{n!}{n^n}}$

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }\frac{n!}{n!}=}{\displaystyle \sum _{n=0}^{\infty }{a}_n}\\ a_n=\frac{n!}{n!}\\ a_{n+1}=\frac{\left(n+1\right)!}{{\left(n+1\right)}^{n+1}}\\ \underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}\\ =\underset{n\to \infty }{\lim }\frac{\left(n+1\right)!\left(n^n\right)}{n!{\left(n+1\right)}^{n+1}}\\ =\underset{n\to \infty }{\lim }\frac{n^n}{{\left(n+1\right)}^n}\\ =\underset{n\to \infty }{\lim }{\left\{\frac{n}{1+n}\right\}}^n\\ \text{Let, }x=\underset{n\to \infty }{\lim }n{\left\{\frac{n}{1+n}\right\}}^n\\ \text{Then }\ln x=\underset{n\to \infty }{\lim }n\ln \left\{\frac{1}{1+\left(1/n\right)}\right\}\\ \ln x=\underset{t\to 0}{\lim }-\frac{\ln \left(1+t\right)}{t}\\ \ln x=-1\\ x=\frac{1}{e}<1\end{array}$