# Calculate characteristic function

$p(n)=(1-r)^2nr^{n-1},n=1,2,...$ $f(z)=1/(1-z)$ has derivative $f'(z)$ with convergent power series $f'(z)=1/(1-z)^2=1+2z+3z^2+...$

the answer I have got is $(1-r)^2e^{it}(1-re^{it})^{-2}$ , I am not sure if it's correct or not.

• @Pierpaolo Vivo – user164945 Feb 9 '16 at 16:30

$$\langle e^{\mathrm{i}t N}\rangle=\sum_{n=1}^\infty (1-r)^2 n r^{n-1}e^{\mathrm{i}t n}=-\mathrm{i}\frac{(1-r)^2}{r}\frac{d}{dt}\sum_{n=1}^\infty (r e^{\mathrm{i}t})^n\ ,$$ and then apply the geometric series $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ (and noting that the term corresponding to $n=0$ does not contribute thanks to the derivative) to get $$\langle e^{\mathrm{i}t N}\rangle=-\mathrm{i}\frac{(1-r)^2}{r}\frac{d}{dt}\frac{1}{1-r e^{\mathrm{i}t}}=\frac{(1-r)^2}{(1-r e^{\mathrm{i}t})^2}e^{\mathrm{i}t}\ .$$