Use the characteristic function of a geometric function to derive $E[X]=\frac{1}{p}$

Question

Suppose $X$ has geometric distribution $Geo(p)$, I want to use the characteristic function $\phi_{X}(t)=\frac{pe^{it}}{1-(1-p)e^{it}}$ to derive $E[X]=\frac{1}{p}$.

I tried $\phi_{X}(t)=E[e^{itx}]=\frac{pe^{it}}{1-(1-p)e^{it}}=\Sigma_{x=0}^{\infty}f_X(x)e^{itx}=\Sigma_{x=0}^{\infty}p(1-p)^{x-1}e^{itx}$ but got stuck from here.

Am I on the right track or is there another way to get $E[X]$ from $E[e^{itx}]$?

Answer 1

$\frac d {dt} \phi_X(t)=iEXe^{itX}$ and evaulating this at $t=0$ gives $iEX$. So calculate the left side using the formula for $\phi_X(t)$; use quotient rule for differentiation.