# How the equation is derived using geometric sequence

I need to know how equation (3.7) is derived from equation (3.6). I contacted the authors and they said "expression inside parenthesis is the summation geometric sequence with ratio of (p1/p2)." In my answer I am getting the same term without addition of 1 in the term. How this 1 comes here. Can someone kindly write the steps. Thanks in advance

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Apparently (3.6) is valid only for $n\ge 1$, so we have
\begin{align*} 1&=\pi_0+\sum_{n\ge 1}\pi_n\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\sum_{n\ge 1}\left(\frac{p_1}{p_2}\right)^{n-1}\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\sum_{n\ge 0}\left(\frac{p_1}{p_2}\right)^n\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\frac1{1-\frac{p_1}{p_2}}\\ &=\pi_0+\pi_0\left(\frac{\alpha}{p_2}\right)\frac{p_2}{p_2-p_1}\\ &=\pi_0\left(1+\frac{\alpha}{p_2}\cdot\frac{p_2}{p_2-p_1}\right)\;, \end{align*}
@Osman: That was my typo, but I’ve fixed it now: $\alpha/p_2$ is correct. – Brian M. Scott Dec 27 '12 at 21:46
@Osman: Just to make sure: (1) I’ve separated the $\pi_0$ term from the rest. (2) Going from $\sum_{n\ge 1}x^{n-1}$ to $\sum_{n\ge 0}x^n$ can be thought of like this: first let $k=n-1$, so that $n=k+1$, and $\sum_{n\ge 1}x^{n-1}=\sum_{k+1\ge 1}x^k=\sum_{k\ge 0}x^k$, and then change the name of the index variable back to $n$. – Brian M. Scott Dec 28 '12 at 7:36