Question

Find the sum of $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+....+\frac{2^n{x}^{2^n-1}}{1+x^{2^n}}$.

Answer 1

Let $f\left(x\right)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+.....+\frac{2^n{x}^{2^n-1}}{1+x^{2^n}}$

We see that in every term of $f\left(x\right)$, the numerator is the derivative of the denominator.

Taking advantage of this, we integrate $f\left(x\right)$

$\Rightarrow \int f\left(x\right)dx=ln(1+x)+ln(1+x^2)+ln(1+x^4)+...+ln(1+x^{2^n})$

$\therefore \int f\left(x\right)dx=ln\left[\right(1+x\left)\right(1+x^2)...(1+x^{2^n}\left)\right]$

Now, we multiply and divide by (1-x) inside the logarithm

$\therefore \int f\left(x\right)dx=ln[\frac{(1-x)(1+x)(1+x^2)...(1+x^{2^n})}{(1-x)}]$

Using $(a+b)(a-b)=a^2-b^2$ repetitively in the numerator, we get

$\therefore \int f\left(x\right)dx=ln\frac{(1-x^{2^{n+1}})}{(1-x)}=ln(1-x^{2^{n+1}})-ln(1-x)$

Now, differentiating on both sides w.r.t. x, we get

$\therefore f\left(x\right)=\frac{1}{(1-x^{2^{n+1}})}(-x^{2^{n+1}-1})+\frac{1}{(1-x)}$

This is the value of the required sum. If limits are asked, remember to put them at the end, after the differentiation.