I'm trying get the value of the summation for basic queueing theory: $$p\sum_{i=2}^∞ (i-1) ϱ^{i}$$
So we take of ϱ out: $$p*ϱ\sum_{i=2}^∞ (i-1) ϱ^{i-1}$$
however I'm not sure what to do after that. When seeing the answer, it involves derivatives. Can someone please explain?
On
$$S=p\sum_{i=0}^\infty (i-1)\rho^i=p\sum_{k=1}^\infty k\rho^{k+1}=p\rho^2\sum_{k=1}^\infty k\rho^{k-1}=p\rho^2\left(\sum_{k=1}^\infty \rho^{k} \right)'$$ $$\sum_{k=1}^\infty \rho^{k}=\frac{\rho }{1-\rho}=\frac{1 }{1-\rho}-1$$ $$\left(\sum_{k=1}^\infty \rho^{k} \right)'=\frac{1}{(1-\rho)^2}$$ $$S=\frac{p\rho^2}{(1-\rho)^2}$$
$$ p\sum_{i=0}^\infty (i-1)\rho^i=p\sum_{k=1}^\infty k\rho^{k+1}$$
Then, assuming $\vert \rho\vert<1$
\begin{eqnarray} \sum_{k=1}^\infty k\rho^{k+1}&=&\rho^2+2\rho^3+3\rho^4+4\rho^5+\cdots\\ &=&\sum_{n=1}^\infty\sum_{k=n}^\infty\rho^{k+1}\\ &=&\sum_{n=1}^\infty\left(\rho^{n+1}\sum_{k=0}^\infty\rho^{k}\right)\\ &=&\sum_{n=1}^\infty\left(\rho^{n+1}\frac{1}{1-\rho}\right)\\ &=&\frac{1}{1-\rho}\sum_{n=1}^\infty\rho^{n+1}\\ &=&\frac{\rho^2}{1-\rho}\sum_{n=0}^\infty\rho^{n}\\ &=&\left(\frac{\rho}{1-\rho}\right)^2 \end{eqnarray}
So we get
$$ p\sum_{i=0}^\infty (i-1)\rho^i=p\left(\frac{\rho}{1-\rho}\right)^2$$
Addendum:
I am adding the following for those having difficulty following the summations.
\begin{eqnarray} \rho^2+2\rho^3+3\rho^4+4\rho^5+\cdots&=&\rho^2+\rho^3+\rho^4+\rho^5+\cdots\\ &+&\rho^3+\rho^4+\rho^5+\rho^6+\cdots\\ &+&\rho^4+\rho^5+\rho^6+\rho^7+\cdots\\ &+&\rho^5+\rho^6+\rho^7+\rho^8+\cdots\\ &\vdots&\\ &=&\rho^2(1+\rho+\rho^2+\rho^3+\cdots)\\ &+&\rho^3(1+\rho+\rho^2+\rho^3+\cdots)\\ &+&\rho^4(1+\rho+\rho^2+\rho^3+\cdots)\\ &+&\rho^5(1+\rho+\rho^2+\rho^3+\cdots)\\ &\vdots&\\ &=&(\rho^2+\rho^3+\rho^4+\rho^5+\cdots)(1+\rho+\rho^2+\rho^3+\cdots)\\ &=&\rho^2(1+\rho+\rho^2+\rho^3+\cdots)(1+\rho+\rho^2+\rho^3+\cdots)\\ &=&\rho^2\left(\frac{1}{1-\rho}\right)\left(\frac{1}{1-\rho}\right)\\ &=&\left(\frac{\rho}{1-\rho}\right)^2 \end{eqnarray}