Expert: Paul Klarreich - 2/1/2009

Question
QUESTION: Right now I am learning about Infinite Geometric Series in
one of my classes and I've read everything I can about this
stuff but I'm still not quite sure how to solve this
problem. The problem is

Infinity
(The Summation Notation Sym.      i/(4^i)
I don't know how to insert it.)
i = 0

And I have to find the sum. I know you have the common
ratio but the numbers I am getting just don't seem right.
Any insight would be greatly appreciated!

ANSWER: Questioner:   Jake
Category:  Advanced Math
Private:  No
 
Subject:  Infinite Geometric Series
Question:  Right now I am learning about Infinite Geometric Series in
one of my classes and I've read everything I can about this
stuff but I'm still not quite sure how to solve this
problem. The problem is

Infinity
(The Summation Notation Sym.      i/(4^i)
I don't know how to insert it.)
i = 0

And I have to find the sum. I know you have the common
ratio but the numbers I am getting just don't seem right.
Any insight would be greatly appreciated!
.........................................
Hi, Jake,

Your expression seems to be:

SUM[I=0 TO infinity] i/(4^i)   << use this notation.

which looks like this: (the i = 0 term is zero)

1     2     3     4     5
--- + --- + --- + --- + --- + ...
4^1   4^2   4^3   4^4   4^5

which is NOT a G.S. because the numerators are not constants.

Did you mean to write:

SUM[I=0 TO infinity] 1/(4^i)   << 1, not 'i'

If so, let me know.  Resend the question and I'll see what I can do.


---------- FOLLOW-UP ----------

QUESTION: Right this is what I got when I was trying to work it out.

****
Your expression seems to be:

SUM[I=0 TO infinity] i/(4^i)   << use this notation.

which looks like this: (the i = 0 term is zero)

1     2     3     4     5
--- + --- + --- + --- + --- + ...
4^1   4^2   4^3   4^4   4^5
*****

I can't seem to get a common ratio for this so does that
mean it can't be summed? And yes I meant to write "i". And
thanks for the help.

Answer
Questioner:   Jake
Category:  Advanced Math
Private:  No
 
Subject:  Infinite Geometric Series
Question:  Right now I am learning about Infinite Geometric Series in
one of my classes and I've read everything I can about this
stuff but I'm still not quite sure how to solve this
problem. The problem is

Infinity
(The Summation Notation Sym.      i/(4^i)
I don't know how to insert it.)
i = 0

And I have to find the sum. I know you have the common
ratio but the numbers I am getting just don't seem right.
Any insight would be greatly appreciated!
.........................................
Hi, Jake,

Your expression seems to be:


SUM[I=0 TO infinity] i/(4^i)

which looks like this: (the i = 0 term is zero)

1     2     3     4     5
--- + --- + --- + --- + --- + ...
4    4^2   4^3   4^4   4^5

which is NOT a G.S. because the numerators are not constants.

Did you mean to write:

SUM[I=0 TO infinity] 1/(4^i)   << 1, not 'i'

If so, let me know.  Resend the question and I'll see what I can do.

=================================================

Questioner:   Jake
Category:  Advanced Math
Private:  No
 
Subject:  Infinite Geometric Series
Question:  QUESTION: Right now I am learning about Infinite Geometric Series in
one of my classes and I've read everything I can about this
stuff but I'm still not quite sure how to solve this
problem. The problem is

Infinity
(The Summation Notation Sym.      i/(4^i)
I don't know how to insert it.)
i = 0

And I have to find the sum. I know you have the common
ratio but the numbers I am getting just don't seem right.
Any insight would be greatly appreciated!

ANSWER: Questioner:   Jake
Category:  Advanced Math
Private:  No

Subject:  Infinite Geometric Series
Question:  Right now I am learning about Infinite Geometric Series in
one of my classes and I've read everything I can about this
stuff but I'm still not quite sure how to solve this
problem. The problem is

Infinity
(The Summation Notation Sym.      i/(4^i)
I don't know how to insert it.)
i = 0

And I have to find the sum. I know you have the common
ratio but the numbers I am getting just don't seem right.
Any insight would be greatly appreciated!
.........................................
Hi, Jake,

Your expression seems to be:

SUM[I=0 TO infinity] i/(4^i)   << use this notation.

which looks like this: (the i = 0 term is zero)

1     2     3     4     5
--- + --- + --- + --- + --- + ...
4^1   4^2   4^3   4^4   4^5

which is NOT a G.S. because the numerators are not constants.

Did you mean to write:

SUM[I=0 TO infinity] 1/(4^i)   << 1, not 'i'

If so, let me know.  Resend the question and I'll see what I can do.


---------- FOLLOW-UP ----------

QUESTION: Right this is what I got when I was trying to work it out.

****
Your expression seems to be:

SUM[I=0 TO infinity] i/(4^i)   << use this notation.

which looks like this: (the i = 0 term is zero)

1     2     3     4     5
--- + --- + --- + --- + --- + ...
4^1   4^2   4^3   4^4   4^5
*****

I can't seem to get a common ratio for this so does that
mean it can't be summed? And yes I meant to write "i". And
thanks for the help.
.................................
Hi, Jake,

I had to be sure, since the calculation is not easy, and if it wasn't what you meant, it would be a waste.
.............
OK, then, it is not a G.S.  HOWEVER, I think we can pull a couple of fancy tricks to sum it:

Call this series P.  (We could call it Shirley, but....)

[And I am going to use 'k', not 'i' -- easier to see.]

P =

1     2     3     4     5
--- + --- + --- + --- + --- + ... =
4^1   4^2   4^3   4^4   4^5

EQUALS......

1     1     1     1     1
--- + --- + --- + --- + --- + ... +  << split off 1/4^k, a G.S.
4^1   4^2   4^3   4^4   4^5

      1     2     3     4   
     --- + --- + --- + --- + ...    
     4^2   4^3   4^4   4^5  


EQUALS..........

1     1     1     1     1
--- + --- + --- + --- + --- + ... +
4^1   4^2   4^3   4^4   4^5

1    1     2     3     4     5
- ( --- + --- + --- + --- + --- + ...)  =  << factor  1/4
4   4^1   4^2   4^3   4^4   4^5

EQUALS...........

1     1     1     1     1
--- + --- + --- + --- + --- + ... +
4^1   4^2   4^3   4^4   4^5

1
- (P)  =  << observe that the thing inside is our original series, P.
4


Now the first part of that THING is:

1     1     1     1     1
--- + --- + --- + --- + --- + ...
4^1   4^2   4^3   4^4   4^5

which IS a G.S. and we can sum it, with  r = 1/4 and  a = 1/4.

   a       1/4       1
= ----- = ------- = ------ = 1/3
 1 - r   1 - 1/4   4 - 1

So we are in business.

P = 1/3 + (1/4)P

3P/4 = 1/3, and  P = 4/9.

A NOTE:  You can probably put the stuff into an Excel worksheet and get a rough idea of the answer and probably a check on your final calculation.