How can I prove that:

$$\sum_{m\geq n}^\infty \frac{(\lambda (1-p)t)^m}{(m-n)!}=((1-p)\lambda t)^n e^{(1-p)\lambda t}$$

  • $\begingroup$ The left side not even defined. $\endgroup$ – Kavi Rama Murthy Nov 13 '18 at 9:07
  • $\begingroup$ @KaviRamaMurthy Better now? $\endgroup$ – Dole Nov 13 '18 at 9:18
  • $\begingroup$ Perhaps the denominator is $(m - n)!$. $\endgroup$ – Awe Kumar Jha Nov 13 '18 at 9:28
  • $\begingroup$ Is $n$ a free choice, or should you be considering $\sum_n \sum_{m \geq n} \ldots$? $\endgroup$ – Kevin Nov 13 '18 at 9:28

I am interpreting $m-n!$ as $(m-n)!$. Just change the variable from $m$ to $j=m-n$. Let $x=\lambda (1-p)t$. You will get $\sum_{j=1}^{\infty} \frac {x^{j+n}} {j!}$. Pull out $x^{n}$ and use the series for $e^{x}$.


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