Chapter 8.4, Problem 32E

To determine

To check: The series is convergent or divergent.

Answer to Problem 32E

  ${\displaystyle \sum _{n=1}^{\infty }{(-2)}^n\cdot \frac{n!}{(2n)!}}$ is convergent series .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=1}^{\infty }{(-2)}^n\cdot \frac{n!}{(2n)!}}$

The series also can be written as

  ${\displaystyle \sum _{n=1}^{\infty }{(-2)}^n\cdot \frac{n!}{(2n)!}}={\displaystyle \sum _{n=1}^{\infty }{(-1)}^n\cdot \frac{2^n\cdot n!}{(2n)!}}$ we know that ,

  ${\displaystyle \sum _{n=k}^{\infty }{(-1)}^n{b}_n}\text{ is converges if }{b}_n\ge 0$

And $\underset{n\to \infty }{\lim }{b}_n=0$ and $b_n$ decreasing sequence .

Also $b_n$ should be decreasing sequence.

Here , $b_n=\frac{2^n\cdot n!}{(2n)!}$

  $\underset{n\to \infty }{\lim }\frac{2^n\cdot n!}{(2n)!}=0$

Hence, ${\displaystyle \sum _{n=1}^{\infty }{(-2)}^n\cdot \frac{n!}{(2n)!}}$ is convergent series .