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Compute the value of  such that $\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\cdots\right)\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots\right)=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\cdots$
 

Guest Jun 24, 2018
 #1
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Can somebody translate this into proper LaTex, please? Thank you.

Guest Jun 24, 2018
 #2
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\($\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\cdots\right)\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots\right)=1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\cdots$\)

 

The sum of the first series on the left hand is given by

 

 1 / (1 - 1/2)  =  1  / (1/2)  = 2

 

And the sum of the second series on the left hand side is given by :

 

1  / ( 1 - ( - 1/2) )  =  1 / (3/2)   = 2/3

 

So.....the  product of these series sums  =    2 * 2/3   = 4/3

 

 

And the common  difference for the series  on the right side is  1/x

 

So....the sum of the  series on the right side  is  given by :

 

     1 / [ 1  - 1/x]  =  1  / [ (x - 1) / x  ]  =    x  / x -1

 

So....setting these equal and solving for x, we have

 

4/3  =   x /  ( x  -1)        cross-multiply

 

4 ( x - 1)  =  3x

 

4x  - 4   = 3x

 

x  =  4

 

 

cool cool cool

CPhill  Jun 25, 2018

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