Question

Calculus3 - Infinite sequences and series ( URGENT!!)​

Answer 1

Limit=0

Converges

Absolutely converges

Explanation:

If
`a_n=(2^n n!)/((3n+4)!)`

then
`a_(n+1)=(2^(n+1) (n+1)!)/((3(n+1)+4)!)`.

Let's rewrite
`a__(n+1)` a little.

I'm going to hone in on (3(n+1)+4)! for a bit.

Distribute: (3n+3+4)!

Combine like terms (3n+7)!

I know when I have to find the limit of that ratio I'm going to have to rewrite this a little more so I'm going to do that here. Notice the factor (3n+4)! in
`a_n`. Some of the factors of this factor will cancel with some if the factors of (3n+7)!

(3n+7)! can be rewritten as (3n+7)×(3n+6)×(3n+5)×(3n+4)!

Let's go ahead and put our ratio together.

`a_(n+1)×(1)/(a_n)`

The second factor in this just means reciprocal of
`{a_n}`.

Insert substitutions:

`(2^(n+1) (n+1)!)/((3(n+1)+4)!)×((3n+4)!)/(2^nn!)`

Use the rewrite for (3(n+1)+4)!:

`(2^(n+1) (n+1)!)/((3n+7)(3n+6)(3n+5)(3n+4)!)×((3n+4)!)/(2^nn!)`

Let's go ahead and cancel the (3n+4)!:

`(2^(n+1) (n+1)!)/((3n+7)(3n+6)(3n+5))×(1)/(2^nn!)`

Use 2^(n+1)=2^n × 2 with goal to cancel the 2^n factor on top and bottom:

`(2^(n)2(n+1)!)/((3n+7)(3n+6)(3n+5))×(1)/(2^nn!)`

`(2(n+1)!)/((3n+7)(3n+6)(3n+5))×(1)/(n!)`

Use (n+1)!=(n+1)×n! with goal to cancel the n! factor on top and bottom:

`(2(n+1)×n!)/((3n+7)(3n+6)(3n+5))×(1)/(n!)`

`(2(n+1))/((3n+7)(3n+6)(3n+5))×(1)/(1)`

Now since n approaches infinity and the degree of top=1 and the degree of bottom is 3 and 1<3, the limit approaches 0.

This means it absolutely converges and therefore converges.