# Simplifying a summation?

While studying Stochastic Process, I came across the following equation which I am trying to reduce it int simple terms!

It is

$$P_{0}\left(1 + \sum_{n=1}^{(s-1)} \frac{\rho^n}{n!} + \sum_{n=s}^{\infty}\frac{\rho ^n}{s! (s^{n-s})}\right) = 1$$

I have to evaluate $P_{0}$ in simplest form.

I tried to see the terms of the summation in order to see whether there is some pattern in it -

Like, consider
$$\left(\rho + \frac{\rho ^2}{2!} + \frac{\rho^3}{3!} + \cdots+\frac{\rho^{(s-1)}}{(s-1)!}\right) + \left(\frac{\rho^s}{s!}+\frac{\rho^{s+1}}{s!} + \frac{\rho^{s+2}}{2s!} + \cdots\right)$$

So we get $$\left(\rho + \frac{\rho ^ 2}{2!}+\frac{\rho ^ 3}{3!}+\cdots\right)+\frac{\rho^s}{s!}\left(1 + \rho + \frac{\rho ^ 2}{2}+\frac{\rho ^ 3}{3}+\cdots\right)$$

I am not seeing any pattern so that I can simplify it?

• For second term i got,$(1 + \rho + \frac{\rho ^2}{2}+...) = ln(1/(1-\rho))$ – BAYMAX Apr 10 '18 at 8:09
• But If now I could get a simplified form of $\sum_{n=1}^{s-1}\frac{\rho^n}{n!}$ ? then it would be simplified!! – BAYMAX Apr 10 '18 at 8:13
• is it just me or does the expansion of $1+ \sum_{n=1}^{(s-1)}\frac{\rho^n}{n!}$ look similar to the expansion of $e^x$ ? – The Integrator Apr 10 '18 at 8:20
• would you be so kind as to show how you for $\ln(\frac1{1-\rho})$. I wasnt able to get it. – The Integrator Apr 10 '18 at 8:23
• it looks exponential if it would have been infinite series right also i got the summation for $ln$ from wiki - en.wikipedia.org/wiki/Natural_logarithm , in which it had shown the expansion of $ln(\frac{x}{x-1})$ – BAYMAX Apr 10 '18 at 8:26

$A=1 + \sum\limits_{n=1}^{s-1} \frac{\rho^n}{n!} + \sum\limits_{n=s}^{\infty}\frac{\rho ^n}{s! (s^{n-s})}=\sum\limits_{n=0}^{s-1} \frac{\rho^n}{n!}+\sum\limits_{n=s}^{\infty}{(\frac{\rho}{s})}^n \frac{s^s}{s!}$

The last sum is equal to:

$\sum\limits_{n=s}^{\infty}{(\frac{\rho}{s})}^n \frac{s^s}{s!}=\sum\limits_{n=0}^{\infty}{(\frac{\rho}{s})}^n \frac{s^s}{s!}-\sum\limits_{n=0}^{s-1}{(\frac{\rho}{s})}^n \frac{s^s}{s!}$

If the absolute value of $\rho\lt {s}$ then we can write:

$\frac{s^s}{s!}\big(\frac{1}{1-\frac{\rho}{s}} - \frac{1-{(\frac{\rho}{s})}^s}{1-\frac{\rho}{s}}\big)=\frac{\rho^s}{s!}\frac{s}{s-p}$

Turn back to the starting expression we get:

$P\big(A\big)=P\big(\sum\limits_{n=0}^{s-1} \frac{\rho^n}{n!}+\frac{\rho^s}{s!}\frac{s}{s-p}\big)$

If $s\rightarrow \infty$ then $A \rightarrow {e}^\rho$