Question

I really need help with this practice problem It asks to answer (a) and (b) Please put these separately so know which is which ^

Answer 1

anGiven the infinite series:

`\sum ^(\infty)_(n\mathop=1)((2n!)/(2^(2n)))`

You need to remember that, by definition, given an infinite series:

`\sum ^(\infty)_(n\mathop=1)a_n`

(a) The formula for applying the Ratio Test is:

`\lim _(n\rightarrow\infty)(|a_(n+1)|)/(|a_n|)=L`

By definition:

1. If:

`L<1`

The series converges.

2. If:

`L>1`

Or:

`L=\infty`

The series diverges.

3. If:

`L=1`

The Ratio Test is inconclusive.

Therefore, you need to set up:

`\lim _(n\rightarrow\infty)\frac{2(n+1)!}{2^(2(n+1))_{}}\cdot\frac{2^(2n)_{}}{2n!}`

Simplifying, you get:

`\lim _(n\rightarrow\infty)\frac{2(n+1)!}{2^(2(n+1))_{}}\cdot\frac{2^(2n)_{}}{2n!}=((n+1)!)/(n!)\cdot(2^(2n))/(2^(2(n+1)))=(n+1)\cdot2^(2n-2n-2)=(n+1)\cdot2^(-2)=\infty`

(b) Notice that:

`r=\infty`

Therefore, this indicates that the series diverges.

Hence, the answers are:

(a)

`r=\infty`

(b) It tells that the series diverges.