Question

If $x<1$, then $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots \infty =$

• A. $-\frac{1}{1+x}$
• B. $\frac{1}{1-x}$
• C. $-\frac{1}{1-x}$
• D. $\frac{1}{1+x}$

Answer 1

The correct option is B $\frac{1}{1-x}$

The given series is in the form
$\frac{{f_1}^{\prime }\left(x\right)}{f_1\left(x\right)}+\frac{{f_2}^{\prime }\left(x\right)}{f_2\left(x\right)}+\frac{{f_3}^{\prime }\left(x\right)}{f_3\left(x\right)}+\cdots \infty$
Then consider the product $f_1\left(x\right).f_2\left(x\right).f_3\left(x\right)\cdots f_n\left(x\right)$. Now,
$(1-x)(1+x)(1+x^2)(1+x^4)\cdots (1+x^{2^{n-1}})$ ---------(1)
$=(1-x^2)(1+x^2)(1+x^4)\cdots (1+x^{2^{n-1}})$
$=(1-x^4)(1+x^4)\cdots (1+x^{2^{n-1}})$
$⋮$
$=(1-x^{2^{n-1}})(1+x^{2^{n-1}})$
$=1-x^{2^n}$
Now, when $n\to \infty$, $x^{2^{n-1}}\to 0(\because x<1)$.
Therefore, taking $n\to \infty$, in (1), we get
$(1-x)(1+x)(1+x^2)(1+x^4)\cdots =1$
Taking logarithm, we get
$\log (1-x)+\log (1+x)+\log (1+x^2)+\log (1+x^4)+\cdots =0$
Differentiating w.r.t. $x$, we get
$-\frac{1}{1-x}+\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots =0$
or $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots \infty =\frac{1}{1-x}$