How can i express this Infinite series without a summation symbol?

  • #1
bobthebanana
23
0
(1/2) + (2/4) + ... + (n/(2^n))

=

sum i=1 to i=infinity of (i/(2^i))?


i know how to express the sum of just 1/(2^i), but not the above

thanks for the help!
 

Answers and Replies

  • #2
arildno
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I don't understand the question.
1. Do you want a symbolic notation for a sum without using the standard summation symbol?
2. Are you instead asking for how the standard summation symbol looks like?
3. Do you wish to find an alternate expression for the sum, i.e, calculate it in a manner so that it is easy to evaluate it for an arbitrarily chosen n?
 
  • #3
arildno
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I think you mean 3, i.e, how to calculate that sum. Am I right?
 
  • #4
bobthebanana
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yea number 3
 
  • #5
ice109
1,714
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[tex]2^{-n} \left(-n+2^{n+1}-2\right)[/tex]
 
  • #6
Pere Callahan
586
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[tex]2^{-n} \left(-n+2^{n+1}-2\right)[/tex]

This is so instructive...!:rolleyes:
 
  • #7
exk
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the summation is not 0, the above is only slightly correct ;)
 
  • #8
ice109
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  • #9
arildno
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Let us consider the function:
[tex]F(x)=\sum_{i=1}^{\infty}i*x^{i}[/tex]
Note that F(1/2) equals your sum!
Now, we may write:
[tex]F(x)=x*\sum_{i=1}^{\infty}i*x^{i-1}=x*\frac{d}{dx}\sum_{i=1}^{\infty}x^{i}=x*\frac{d}{dx}(\frac{1}{1-x}-1)=\frac{x}{(1-x)^{2}}[/tex]
Hence, we easily gain F(1/2)=2.

For arbitrary finite n, use a similar procedure.
 
  • #10
LukeD
355
3
Bah, I wrote a similar response using generating functions... twice... and physics forums died on me both times so nothing was posted.

Arildno's response is entirely correct though
 

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