Determine whether the following series converes or diverges and find the sum

[kuczmama]

Homework Statement

infinity
Ʃ(2ⁿ+3ⁿ)/(4ⁿ⁺¹)
n=0

Homework Equations

We learned the integral test. The p-series. The nth term test.

The Attempt at a Solution

I figured out that the terms are positive and that they approach 0. the first couple of terms are (2/4)+ 5/16+13/64+35/256+...

I think I can use the integral test, but that only tells me whether or not it converges. I think that it does. However, I don't know how to find its sum. If you guys could help me out I would really appreciate it.

 

[LCKurtz]

Think about geometric series.

 

[kuczmama]

I know that a geometric series is like a+ar+ar²+ar³+...arⁿ, but I don't know how to re-write this in that form

 

[LCKurtz]

I know that a geometric series is like a+ar+ar²+ar³+...arⁿ, but I don't know how to re-write this in that form

Break it up into two problems and factor a 4 out of the denominator.

 

[kuczmama]

do you mean break it up into two integrals? or would it be like

Ʃ2ⁿ/4ⁿ⁺¹ +Ʃ 3ⁿ/4ⁿ⁺¹

Can you break it up like that?

 

[kuczmama]

I can't find a common r value. I don't think it is a geometric series. So I am pretty lost still.

 

[LCKurtz]

Break it up into two problems and factor a 4 out of the denominator.

do you mean break it up into two integrals? or would it be like

Ʃ2ⁿ/4ⁿ⁺¹ +Ʃ 3ⁿ/4ⁿ⁺¹

Can you break it up like that?

I can't find a common r value. I don't think it is a geometric series. So I am pretty lost still.

Yes you can break it up like that. It makes two separate problems. What about my other suggestion in red above?

 

[kuczmama]

this is probably a dumb question, but how do I factor something out if it is a sequence. I really don't know how to factor a 4 out of the denominator. like would I multiply the problem by 1/4

 

[LCKurtz]

$$\frac 1 {4^{n+1}} = \frac 1 4\cdot\frac 1 {4^n}$$

 

[kuczmama]

Ohh ok. Thank you so much. I just figured it out. I now know that the sum is (3/2). I didnt know you could factor it out like that. that helps so much.