Proving $ \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^2}$ for $z \in (-1, 1)$

Question

I do not know where to start, any hints are welcome.

Answer 1

Without differentiation

$$\sum_{n=1}^\infty nz^{n-1} = 1+2z+3z^2+\cdots \tag 1$$

$$\sum_{n=1}^\infty nz^{n} = z+2z^2+3z^3+\cdots \tag 2$$

Subtract (2) from (1)

$$\sum_{n=1}^\infty nz^{n-1}(1-z) = 1+z+z^2+\cdots$$

$$\sum_{n=1}^\infty nz^{n}= \frac{z}{(1-z)^2}$$

Answer 2

Note that $$\sum_{n=0}^\infty z^n=\frac{1}{1-z} \tag 1$$

for $|z|<1$.

Differentiating $(1)$ and multiplying by $z$ (this is legitimate since for any $r<1$, $\sum_{n=1}^\infty nz^{n-1}$ converges uniformly for $|z|\le r<1$) yields

$$\begin{align} z \frac{d}{dz}\sum_{n=0}^\infty z^n&=\sum_{n=0}^\infty nz^n\\\\ &=\sum_{n=1}^\infty nz^n\\\\ &=\frac{z}{(1-z)^2} \end{align}$$

for $|z|<1$.

Answer 3

Another way: $$S(z)=z+2z^2+3z^3+4z^4+\cdots\Rightarrow $$ $$zS(z)=z^2+2z^3+3z^4+4z^5+\cdots\Rightarrow$$ $$\Rightarrow S(z)-zS(z)=S(z)(1-z)=z+z^2+z^3+\cdots=-1+(1+z+z^2+z^3+\cdots)$$ $$=-1+\frac{1}{1-z}=\frac{z}{1-z}\Rightarrow S(z)=\frac{z}{(1-z)^2}\Rightarrow \sum_{n=1}^{+\infty}nz^n=\frac{z}{(1-z)^2}\quad (|z|<1).$$

EDIT: I didn't see Zaid Alyafeai answer.