Question

For $x>1$, the sum of the series $\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^4}{(x+1)(x^2+1)(x^4+1)}+\cdots \infty$ is

Answer 1

Let $S_{\infty }=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^4}{(x+1)(x^2+1)(x^4+1)}+\cdots \infty$
$\frac{x}{x+1}=1-\frac{1}{x+1}\frac{x^2}{(x+1)(x^2+1)}=\frac{1}{x+1}-\frac{1}{(x+1)(x^2+1)}\frac{x^4}{(x+1)(x^2+1)(x^4+1)}=\frac{1}{(x+1)(x^2+1)}-\frac{1}{(x+1)(x^2+1)(x^4+1)}$

Similarly, for the $n^{\text{th}}$ term,
$\frac{x^{2^{n-1}}}{(x+1)(x^2+1)\cdots (x^{2^{n-1}}+1)}=\frac{1}{(x+1)(x^2+1)\cdots (x^{2^{n-2}}+1)}-\frac{1}{(x+1)(x^2+1)\cdots (x^{2^{n-1}}+1)}{S}_n=T_1+T_2+T_3+\cdots +T_n=1-\frac{1}{(x+1)(x^2+1)\cdots (x^{2^{n-1}}+1)}$

If denominator of a fraction is very large, then fraction will become $0$
$\therefore S_{\infty }=1$