Question #f639b

2 Answers
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(n/(n+1))^(n).

It is well known that ((n+1)/n)^{n}=(1+1/n)^{n}->e as n-> infty. Therefore, a_{n+1}/a_{n} -> 2/e approx 0.73576<1 as n-> 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)^oo sqrt(2pin)(2/e)^n le sum_(n=1)^oo (n+1)(2/e)^n

and from

sum (k+1)x^k = d/dx(sum x^(k+1)) the convergence for sum (k+1)x^k is assured if abs x < 1 or in our case

2/e < 1 and as a consequence

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