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# Compute the value of such that

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

Jun 24, 2018

### 2+0 Answers

#1
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Can somebody translate this into proper LaTex, please? Thank you.

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

Jun 25, 2018