Answer:
Limit=0
Converges
Absolutely converges
Step-by-step explanation:
If [tex]a_n=\frac{2^n n!}{(3n+4)!}[/tex]
then [tex]a_{n+1}=\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}[/tex].
Let's rewrite [tex]a__{n+1}[/tex] 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 [tex]a_n[/tex]. 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.
[tex]a_{n+1}×\frac{1}{a_n}[/tex]
The second factor in this just means reciprocal of [tex]{a_n}[/tex].
Insert substitutions:
[tex]\frac{2^{n+1} (n+1)!}{(3(n+1)+4)!}×\frac{(3n+4)!}{2^nn!}[/tex]
Use the rewrite for (3(n+1)+4)!:
[tex]\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)(3n+4)!}×\frac{(3n+4)!}{2^nn!}[/tex]
Let's go ahead and cancel the (3n+4)!:
[tex]\frac{2^{n+1} (n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}[/tex]
Use 2^(n+1)=2^n × 2 with goal to cancel the 2^n factor on top and bottom:
[tex]\frac{2^{n}2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{2^nn!}[/tex]
[tex]\frac{2(n+1)!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}[/tex]
Use (n+1)!=(n+1)×n! with goal to cancel the n! factor on top and bottom:
[tex]\frac{2(n+1)×n!}{(3n+7)(3n+6)(3n+5)}×\frac{1}{n!}[/tex]
[tex]\frac{2(n+1)}{(3n+7)(3n+6)(3n+5)}×\frac{1}{1}[/tex]
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.
