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Sum of a Geometric series

  • Thread starter Mathman23
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  • #1
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Hi Folks,

I have this here geometric series which I'm supposed to find the sum of:

Given

[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

I the sum into sub-sums

[tex]\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1} [/tex]

taking [tex]2^{-n}[/tex]

Since [tex]x^n[/tex] converges towards 1/1+x therefore I differentiate on both sides

[tex]1/(1+x)^2 = \sum_{n=0} ^{\infty} n \cdot 2^{-n} x^{n-1}[/tex]

I multiply with x on both sides and obtain

[tex]x/(1+x)^2 = \sum_{n=0} ^{\infty} n * 2^{-n} x^n[/tex]

if I set x = 1 on both sides I get

[tex](1/4) ? = \sum_{n=1} ^{\infty} n*2^{-n} = 2[/tex]

My teacher says that the expression has to give 2 on the left side, and not (1/4).

What am I doing wrong? Any surgestions?

Best Regards
Fred
 
Last edited:

Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
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Your first line is wrong. When you break your sum up into two sums, you don't do it right. Also, when you say xn converges towards 1/1+x, I suppose you mean [itex]\sum x^n[/itex] converges to 1/1+x. However, this is wrong, it converges to 1/1-x. After that, I have no idea what you're doing. I don't know where you're getting the "=" sign from. Sorry, I think just about everything in your post makes no sense.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

is not a geometric series!
 
  • #4
Curious3141
Homework Helper
2,843
86
Mathman23 said:
Hi Folks,

I have this here geometric series which I'm supposed to find the sum of:

Given

[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

I the sum into sub-sums

[tex]\sum_{n=0} ^{\infty} 2^{-n} + \sum_{n=0} ^{\infty} \frac{1}{2}^{n-1} [/tex]

taking [tex]2^{-n}[/tex]

Since [tex]x^n[/tex] converges towards 1/1+x therefore I differentiate on both sides

[tex]1/(1+x)^2 = \sum_{n=0} ^{\infty} n \cdot 2^{-n} x^{n-1}[/tex]

I multiply with x on both sides and obtain

[tex]x/(1+x)^2 = \sum_{n=0} ^{\infty} n * 2^{-n} x^n[/tex]

if I set x = 1 on both sides I get

[tex](1/4) ? = \sum_{n=1} ^{\infty} n*2^{-n} = 2[/tex]

My teacher says that the expression has to give 2 on the left side, and not (1/4).

What am I doing wrong? Any surgestions?

Best Regards
Fred
Firstly, you separated the sum wrongly. Here's the correct version :

[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n} = \sum_{n=0} ^{\infty}(n.2^{1-n})+\sum_{n=0} ^{\infty}(2^{-n})[/tex]

Hint : For the first part-sum on the RHS, try dividing by 2 then subtracting away from the original part-sum.
 
Last edited:
  • #5
Curious3141
Homework Helper
2,843
86
HallsofIvy said:
[tex]\sum_{n=0} ^{\infty} \frac{2n+1}{2^n}[/tex]

is not a geometric series!
Actually, it works out to be one ! :biggrin:
 

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