Find $X_0$ in $X_n = \frac{p^n}{2^{n-1}}\cdot X_o$, with $\sum_{n=0} ^\infty X_n =1$

Question

I have the relation $X_n = \frac{p^n}{2^{n-1}}\cdot X_o$, with $\sum_{n=0} ^\infty X_n =1$. I want to find $X_0$. I got:

$\sum_{n=0} ^\infty X_n = 1 $

$\Longrightarrow \sum_{n=0} ^\infty \frac{p^n}{2^{n-1}}\cdot X_o = 1 $

$\Longrightarrow 2 X_0 \sum_{n=0} ^\infty (\frac{p}{2})^n = 1 $

$\Longrightarrow 2 X_0 \cdot \frac{1}{1- \frac{p}{2}} = 1$

$ \Longrightarrow 2 X_0 = 1- \frac{p}{2}$

$ \Longrightarrow X_0 = \frac{1- \frac{p}{2}}{2} = \frac{2-p}{4}$

But I have a solution that says $X_0 = \frac{1- \frac{p}{2}}{1 + \frac{p}{2}}$. Did I something wrong or is the solution wrong?

Answer 1

You must exclude $X_0$ from summation, indeed $X_n=\frac{p^n}{2^{n-1}}X_0$ for $n\geq 1$, since otherwise you have $2X_0=X_0$ which is true only for $X_0=0$. So your solution is wrong