Question

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

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

Answer 1

Answer: (C)

$\frac{1}{1-x}$

Here, $T_n=\frac{2^n{x}^{2^n-1}}{1+x^{2^n}}$ with $T_0=\frac{1}{1+x}=\frac{1-x}{1+x}$

Let, $S_{n+1}={\sum }_{r=0}^n{T}_r$

$S_1=T_0$

$S_2=\frac{1+x+x^2-3x^3}{1-x^4}$

$S_3=\frac{1+x+x^2+x^3+x^4+x^5+x^6-7x^7}{1-x^8}$

So, $S_n=\frac{1+x+x^2+...+x^{2^n-2}-(2^n-1)x^{2^n-1}}{1-x^{2^n}}$

ie, ${lim}_{n\to \infty }{S}_n={lim}_{n\to \infty }\frac{1+x+x^2+...+x^{2^n-2}}{1-x^{2^n}}-{lim}_{n\to \infty }\frac{(2^n-1)x^{(2^n-1)}}{1-x^{2^n}}=\frac{{lim}_{n\to \infty }1+x+x^2+...+x^{2^n-2}}{{lim}_{n\to \infty }1-x^{2^n}}+0$

$\Rightarrow {lim}_{n\to \infty }{S}_n=1+x+x^2+...=(1-x{)}^{-1}=\frac{1}{1-x}$

So, Option C is the correct Option.