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I saw this problem in a 'classical algebra' textbook:

If $y=2x^2-1$ prove that under certain condition we have:

$\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6}+ . . . =\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5y^5}+ . . .$

My attempt:

$\ln (1-\frac{1}{x^2})=-(\frac{1}{x^2}+\frac{1}{2x^4}+\frac{1}{3x^6} + . . .)$

$2\ln(1+\frac{1}{y})=-(\frac{1}{y^2}+\frac{1}{2y^4}+\frac{1}{3y^6} + . . .) + (\frac{2}{y}+\frac{2}{3y^3}+\frac{2}{5y^5}+ . . .)$

Therefore:

$ \ln (1+\frac {1}{y})^2+\ln(1-\frac{1}{x^2})+\ln(1-\frac{1}{y^2})=0$

$\ln (1+\frac {1}{y})^2(1-\frac{1}{y^2})=\ln(1-\frac{1}{x^2})^{-1}$

$(1+\frac {1}{y})^2(1-\frac{1}{y^2})=(\frac{x^2}{x^2-1})$

Now to find condition should I substitute $y=2x^2-1$ and find a relation for x?

Mourad
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sirous
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1 Answers1

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By What is the correct radius of convergence for $\ln(1+x)$?, we need $$\dfrac1{x^2},\dfrac1{y^2}<1$$

Now,

$$2\sum_{r=0}^\infty\dfrac{(1/y)^{2r+1}}{2r+1}=\ln\dfrac{1+1/y}{1-1/y}=\ln\dfrac{y+1}{y-1}$$

$$\implies-\ln\left(1-\dfrac1{x^2}\right)=\ln\left(\dfrac{y+1}{y-1}\right)=-\ln\left(\dfrac{y-1}{y+1}\right)$$

$$\implies1-\dfrac1{x^2}= \dfrac{y-1}{y+1} $$

Now simplify

lab bhattacharjee
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