Evaluate $\sum_{n=1}^\infty nx^{n-1}$

How can you evaluate $\sum_{n=1}^\infty nx^{n-1} = \frac{1}{(1-x)^2}$ without relying on the fact that it's the derivative of $\sum_{n=1}^\infty x^n = \frac{1}{1-x} $?

Solution 1:

$$a=1+2x+3x^2+4x^3+5x^4+6x^5+...\\|x|<1\\$$multiply a by x $$ xa=x+2x^2+3x^3+4x^4+5x^5+6x^6+...$$now subtract a and ax $$a-xa=1+(2x-x)+(3x^2-2x^2)+(4x^3-3x^3)+...\\a(1-x)=1+x+x^2+x^3+x^4+x^5+...\\ $$$$ a(1-x)=\frac{1}{1-x}\\a=\frac{1}{(1-x)^2}$$

Solution 2:

another way to prove is : $$ a=1+2x+3x^2+4x^3+5x^4+6x^5+...\\$$$$lhs=1+x+x^2+x^3+x^4+x^5+x^6+...\\+x+x^2+x^3+x^4+x^5+x^6+...\\+x^2+x^3+x^4+x^5+x^6+...\\+x^3+x^4+x^5+x^6+...\\x^4+x^5+x^6+...\\...\\$$so$$lhs=1+x+x^2+x^3+x^4+x^5+x^6+...=\frac{1}{1-x}\\+x+x^2+x^3+x^4+x^5+x^6+...=\frac{x}{1-x}\\+x^2+x^3+x^4+x^5+x^6+...=\frac{x^2}{1-x}\\+x^3+x^4+x^5+x^6+...=\frac{x^3}{1-x}\\x^4+x^5+x^6+... =+\frac{x^4}{1-x}\\...\\$$factor denominator $$\frac{1}{1-x}+ \frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\frac{x^4}{1-x}+...\\\frac{1}{1-x}(1+x+x^2+x^3+x^4+x^5+x^6+...)=\frac{1}{1-x}*\frac{1}{1-x}$$