Geometric Series Homework: Sum of ((n+1)*3^n)/2^(2n)

[aselin0331]

Homework Statement

The sum of ((n+1)*3^n)/(2^2n)

Homework Equations

absolute value of r must be less than 1 for the series to be convergent.

The Attempt at a Solution

i tried multiplying it out and splitting it up like:

3^n*n/(2^(2n))+3^n/(2^(2n))

but then i am stuck when I try to pull the nth power out...help?

 

[micromass]

Are you sure you had to do this with geometric series?? Because this isn't a geometric serie, and it can't be brought in the form of a geometric serie...

 

[Mark44]

Homework Statement

The sum of ((n+1)*3^n)/(2^2n)

Homework Equations

absolute value of r must be less than 1 for the series to be convergent.

The Attempt at a Solution

i tried multiplying it out and splitting it up like:

3^n*n/(2^(2n))+3^n/(2^(2n))

I don't see any point in doing this.

but then i am stuck when I try to pull the nth power out...help?

What tests do you know that you can use? Ratio test might be one to try.

 

[gomunkul51]

((n+1)*3^n)/(2^2n) = ((n+1)*3^n)/(4^n) = (n+1)*(3/4)^n

This is not a geometric series because the coefficient of q^n is not constant.

To determine convergence you can use the ratio test or the root test.

 

[aselin0331]

I guess I should have said that this multiple choice question is asking us to select the reason why the series converges and these are the options:

A. Convergent geometric series
B. Convergent p series
C. Comparison(or limit comparison) with a geometric or P-series
D. Converges by alternating series test

I have a feeling that it is limit comparison but when i multiply it by 4^n/3^n (because I am comparing it to 3^n/4^n, it come out to only n+1 and that's infinity.

But this thing converges...I know how to do it with ratio or root which are not options here...

 

[gomunkul51]

If you know it converges (you said so yourself) and the limit comparison test gives you the opposite answer. may be your not using it correctly?

What does the limit comparison test states? also write the full path to your answer.