Prove $\frac{-\log(1-x)}{x(1-x)}=1+(1+1/2)x+(1+1/2+1/3)x^3+...$

Question

Let $0<x<1$. How can i prove the following identity:

$$\frac{-\log(1-x)}{x(1-x)}=1+(1+1/2)x+(1+1/2+1/3)x^3+...\ \ .$$

Answer 1

$$0<x<1 \\1+x+x^2+x^3+x^4+...=\frac{1}{1-x}\\\int(1+x+x^2+x^3+x^4+...)dx=\int \frac{1}{1-x}dx\\x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...=-ln(1-x)\\$$now divide by x $$\frac{-ln(1-x)}{x}=\frac{x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...}{x}=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+...$$now multiply by $1+x+x^2+x^3+x^4+...=\frac{1}{1-x}$ $$\frac{-ln(1-x)}{x}\frac{1}{1-x}=(1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+...)(1+x+x^2+x^3+x^4+...)=\\1+x+\frac{x}{2} +x^2+\frac{x^2}{2}+\frac{x^2}{3}+...$$

Answer 2

In general, let $f(x)=\sum a_nx^n$. Show that $$(1-x)^{-1}f(x)=\sum \sum_{k=0}^n a_k x^n$$