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$p(n)=(1-r)^2nr^{n-1},n=1,2,...$ $f(z)=1/(1-z)$ 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.

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  • $\begingroup$ @Pierpaolo Vivo $\endgroup$ – user164945 Feb 9 '16 at 16:30
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$$ \langle e^{\mathrm{i}t N}\rangle=\sum_{n=1}^\infty (1-r)^2 n r^{n-1}e^{\mathrm{i}t n}=-\mathrm{i}\frac{(1-r)^2}{r}\frac{d}{dt}\sum_{n=1}^\infty (r e^{\mathrm{i}t})^n\ , $$ 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 $$ \langle e^{\mathrm{i}t N}\rangle=-\mathrm{i}\frac{(1-r)^2}{r}\frac{d}{dt}\frac{1}{1-r e^{\mathrm{i}t}}=\frac{(1-r)^2}{(1-r e^{\mathrm{i}t})^2}e^{\mathrm{i}t}\ . $$

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