- #1

bobthebanana

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=

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!

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- Thread starter bobthebanana
- Start date

- #1

bobthebanana

- 23

- 0

=

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!

- #2

arildno

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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

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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...!

- #7

exk

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the summation is not 0, the above is only slightly correct ;)

- #8

ice109

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This is so instructive...!

did he ask for instruction?

- #9

arildno

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[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

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Arildno's response is entirely correct though

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