# Question f639b

Aug 19, 2017

Use the Ratio Test (it's usually a good idea whenever factorials are involved) to show that this series converges.

#### Explanation:

Let a_{n}=(2/n)^{n}*n!.

We want to know whether the series ${\sum}_{n = 1}^{\infty} {a}_{n}$ converges or not.

To apply the Ratio Test, we consider the sequence $| {a}_{n + 1} / {a}_{n} |$. In this example, all the terms in the original sum are positive, so we may dispense with the absolute value signs and note that a_{n+1}/a_{n}=(2/(n+1))^(n+1) * (n+1)!*(n/2)^(n)*1/(n!).

This expression simplifies to:

${a}_{n + 1} / {a}_{n} = 2 {\left(\frac{n}{n + 1}\right)}^{n}$.

It is well known that ${\left(\frac{n + 1}{n}\right)}^{n} = {\left(1 + \frac{1}{n}\right)}^{n} \to e$ as $n \to \infty$. Therefore, ${a}_{n + 1} / {a}_{n} \to \frac{2}{e} \approx 0.73576 < 1$ as $n \to \infty$.

By the Ratio Test, this is enough to imply that the series sum_{n=1}^{infty}(2/n)^{n}*n! converges.

Aug 19, 2017

See below.

#### Explanation:

Using the Stirling asymptotic formula

n! approx sqrt(2pin)(n/e)^n we have

sum_(n=1)^oo(2/n)^2n! approx sum_(n=1)^oo(2/n)^2sqrt(2pin)(n/e)^n = sum_(n=1)^oo sqrt(2pin)(2/e)^n and

${\sum}_{n = 1}^{\infty} \sqrt{2 \pi n} {\left(\frac{2}{e}\right)}^{n} \le {\sum}_{n = 1}^{\infty} \left(n + 1\right) {\left(\frac{2}{e}\right)}^{n}$

and from

$\sum \left(k + 1\right) {x}^{k} = \frac{d}{\mathrm{dx}} \left(\sum {x}^{k + 1}\right)$ the convergence for $\sum \left(k + 1\right) {x}^{k}$ is assured if $\left\mid x \right\mid < 1$ or in our case

$\frac{2}{e} < 1$ and as a consequence

sum_(n=1)^oo(2/n)^2n! # converges.