Question

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

Answer 1

Answer:

Limit=0

Converges

Absolutely converges

Step-by-step explanation:

If $a_n=\frac{2^n n!}{(3n+4)!}$

then $a_{n+1}=\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}$.

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.

$a_{n+1}×\frac{1}{a_n}$

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

Insert substitutions:

$\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}×\frac{(3n+4)!}{2^nn!}$

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

$\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)(3n+4)!}×\frac{(3n+4)!}{2^nn!}$

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

$\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}$

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

$\frac{2^{n}2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}$

$\frac{2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}$

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

$\frac{2(n+1)×n!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}$

$\frac{2(n+1)}{(3n+7)(3n+6)(3n+5)}×\frac{1}{1}$

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.