# Proof that for infinite sum: $\sum_{m\geq n}^\infty \frac{(\lambda (1-p)t)^m}{(m-n)!}=((1-p)\lambda t)^n e^{(1-p)\lambda t}$

How can I prove that:

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

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