Question

Find the sum of the following series $\frac{x}{1.2}+\frac{x^2}{2.3}+\frac{x^3}{3.4}+....$

Answer 1

$\Rightarrow t_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

$\therefore u_n=\frac{x^n}{n}-\frac{x^n}{n+1}$

$u_{n-1}=\frac{x^{n-1}}{n-1}-\frac{x^{n-1}}{n}$

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$u_2=\frac{x^2}{2}-\frac{x^2}{3}$

$u_1=\frac{x}{1}-\frac{x}{2}$

$\therefore S_{\infty }=(\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+.....)-\frac{1}{2}(\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+....)$

$=-\log (1-x)-\frac{1}{2}\{-\log (1-x)-x\}$

$=\frac{1-x}{x}\log (1-x)+1$