Chapter 8, Problem 8RQ

To determine

To find: The statement is true or false and if the statement then given the explanation.

Answer to Problem 8RQ

Theratio test is used to determine that ${\displaystyle \sum \frac{1}{n!}}$ is convergent and the given statement is true. This is proved aslimit $\underset{n\to \infty }{\lim }\left|\frac{\frac{1}{\left(n+1\right)!}}{\frac{1}{n}}\right|$ is equal to 0 and less than 1, the series ${\displaystyle \sum \frac{1}{n!}}$ converges

Explanation of Solution

Given:

The ratio test can be used to determine whether ${\displaystyle \sum \frac{1}{n!}}$ converges.

Calculation:

Consider that the ratio test says that the $\underset{n\to \infty }{\text{lim}}\left|\frac{a_{n+1}}{a_n}\right|<1$ , the series ${\displaystyle \sum _{n\to \infty }^{\infty }{a}_n}$ converges.

Consider if $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|>1$ , then the series diverges and for the limit $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=1$ there is no solution.

Then, for the limit of the ratio of the successive terms is,

  $\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{1}{\left(n+1\right)!}}{\frac{1}{n}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{1}{\left(n+1\right)}\right|\\ =0\end{array}$

Since, limit $\underset{n\to \infty }{\lim }\left|\frac{\frac{1}{\left(n+1\right)!}}{\frac{1}{n}}\right|$ is equal to 0 and less than 1, the series ${\displaystyle \sum \frac{1}{n!}}$ converges.

Thus, the ratio test is used to determine that ${\displaystyle \sum \frac{1}{n!}}$ is convergent and the given statement is true.