Question

If $x<1$, then find the value of $\frac{1-2x}{1-x+x^2}$ $+\frac{2x-4x^3}{1-x^2+x^4}+\frac{4x^3-8x^7}{1-x^4+x^8}+\dots \infty$ when $x=1$

Answer 1

Answer: 1

Here, we have $|x|<1$
Consider
$\left(1+x+x^2\right)\left(1-x+x^2\right)={\left(1+x^2\right)}^2-x^2$
$=1+x^2+x^4$
$\left(1+x+x^2\right)\left(1-x+x^2\right)\left(1-x^2+x^4\right)$
$=\left(1+x^2+x^4\right)\left(1+x^2-x^4\right)$
$={\left(1+x^4\right)}^2-x^4$
$=1+x^4+x^8$
Continuing in the same way, we obtain
$\left(1+x+x^2\right)\left(1-x+x^2\right)\left(1-x^2+x^4\right)$
$...\left(1-x^{2n-1}+x^{2n}\right)$
$=1+x^{2n}+x^{2n+1}$
Taking limit $n\to \infty ,$ we get
$\left(1+x+x^2\right)\left(1-x+x^2\right)$
$\left(1-x^2+x^4\right)\dots =1[$ as $|x|<1]$
Taking log on both sides, we get
$\log \left(1+x+x^2\right)+\log \left(1-x+x^2\right)+\log \left(1-x^2+x^4\right)+\dots =0$
Differentiating both sides w.r.t. $x_1$ we get
$\frac{1+2x}{1+x+x^2}+\frac{-1+2x}{1-x+x^2}+\frac{-2x+4x^3}{1-x^2+xy}+\dots =0$
$\Rightarrow \frac{1-2x}{1-x+x^2}+\frac{2x-4x^3}{1-x^2+x^4}+\dots =\frac{1+2x}{1+x+x^2}$
When $x=1$ , then $\frac{1+2x}{1+x+x^2}=1$