Question

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

A
11+x
B
11x
C
11x
D
11+x

Solution

## The correct option is B $$\displaystyle\frac{1}{1-x}$$The given series is in the form$$\displaystyle\frac{{f_1}^\prime(x)}{{f_1}(x)}+\frac{{f_2}^\prime(x)}{{f_2}(x)}+\frac{{f_3}^\prime(x)}{{f_3}(x)}+\cdots\infty$$Then consider the product $${f_1}(x).{f_2}(x).{f_3}(x)\cdots{f_n}(x)$$. Now,$$(1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{ n-1}})$$ ---------(1)$$=(1-x^2)(1+x^2)(1+x^4)\cdots(1+x^{2^{ n-1}})$$$$=(1-x^4)(1+x^4)\cdots(1+x^{2^{ n-1}})$$$$\:\vdots$$$$=(1-x^{2^{n-1}})(1+x^{2^{ n-1}})$$$$=1-x^{2^{ n}}$$Now, when $$n\rightarrow\infty$$, $$x^{2^{ n-1}}\rightarrow 0\:(\because x<1)$$.Therefore, taking $$n\rightarrow\infty$$, 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$$\displaystyle -\frac{1}{1-x}+\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots=0$$or $$\displaystyle\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\cdots\infty=\frac{1}{1-x}$$Mathematics

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