Proof of an infinite series formula

Question

If $$|x| < 1$$ Prove that

$$\begin{align}\large 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(1 - x)^2}\end{align}$$

Answer 1

Differentiate the geometric series

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

This yields directly $$\frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$

Answer 2

HINT: Let $S= 1+2x+3x^2+4x^3+\cdots$. Then $$S-Sx=1+(2x-x)+(3x^2-2x^2)+\cdots=1+x+x^2+\cdots.$$

Answer 3

Another$^2$ approach:

Since $\sum_{i=0}^{\infty} x^i = \frac1{1-x} $, differentiating both sides we get $\sum_{i=1}^{\infty} ix^{i-1} = \frac1{(1-x)^2} $.

Answer 4

Another approach:

As @Andre said, you have $$S=\sum_{i=0}^\infty x^i=\frac1{1-x}.$$ You wanted to calculate $$T=\sum_{i=1}^\infty ix^{i-1}.$$

Final job: Consider the difference $$xT-S$$ and use the explicit form of $S$.