Question

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

Answer 1

Here, we have $\left|\right.x\left|\right. < 1$
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
$\left(1 + x + x^{2}\right)\left(1 - x + x^{2}\right)\left(1 - x^{2} + x^{4}\right)$
$...\left(1 - x^{2 n - 1} + x^{2 n}\right)$
$=1+x^{2 n}+x^{2 n + 1}$
Taking limit $n \rightarrow \infty ,$ we get
$\left(1 + x + x^{2}\right)\left(1 - x + x^{2}\right)$
$\left(1 - x^{2} + x^{4}\right)\ldots =1\left[\right.$ as $\left|\right.x\left|\right. < 1\left]\right.$
Taking log on both sides, we get
$\log\left(1 + x + x^{2}\right)+\log\left(1 - x + x^{2}\right)+\log\left(1 - x^{2} + x^{4}\right)+\ldots =0$
Differentiating both sides w.r.t. $x_{1}$ we get
$\frac{1 + 2 x}{1 + x + x^{2}}+\frac{- 1 + 2 x}{1 - x + x^{2}}+\frac{- 2 x + 4 x^{3}}{1 - x^{2} + x y}+\ldots =0$
$\Rightarrow \frac{1 - 2 x}{1 - x + x^{2}}+\frac{2 x - 4 x^{3}}{1 - x^{2} + x^{4}}+\ldots =\frac{1 + 2 x}{1 + x + x^{2}}$
When $x=1$ , then $\frac{1 + 2 x}{1 + x + x^{2}}=1$