Question

Prove
$\sum _{i=0}^n{e}^{-\lambda }\frac{{\lambda }^i}{i!}=\frac{1}{n!}{\int }_{\lambda }^{\infty }{e}^{-x}{x}^{n}dx$

Hint: Use integration by parts.

Short answer

The idea of the proob is integrating by parts the right side of the equation $n$times.

Step-by-step solution

Given information

$\sum _{i=0}^n{e}^{-\lambda }\frac{{\lambda }^i}{i!}=\frac{1}{n!}{\int }_{\lambda }^{\infty }{e}^{-x}{x}^{n}dx$

Calculation

The idea is to integrate by parts $n$times

$\frac{1}{n!}{\int }_{\lambda }^{\infty }{e}^{-x}{x}^n\mathrm{d}x=\frac{1}{n!}\left[-{\left.x^n{e}^{-x}\right|}_{\lambda }^{\infty }+n{\int }_{\lambda }^{\infty }{e}^{-x}{x}^{n-1}\mathrm{d}x\right]$

$=\frac{1}{n!}\left[-{\lambda }^n{e}^{-\lambda }+n{\int }_{\lambda }^{\infty }{e}^{-x}{x}^{n-1}\mathrm{d}x\right]$

$=\frac{1}{n!}\left[{\lambda }^n{e}^{-\lambda }+n{\lambda }^{n-1}{e}^{-\lambda }+n(n-1){\int }_{\lambda }^{\infty }{e}^{-x}{x}^{n-2}\mathrm{d}x\right]$

$⋮$

$=\frac{1}{n!}\left[{\lambda }^n{e}^{-\lambda }+n{\lambda }^{n-1}{e}^{-\lambda }+n(n-1){\lambda }^{n-2}{e}^{-\lambda }+\cdots +n!e^{-\lambda }\right]$

$=\frac{e^{-\lambda }{\lambda }^n}{n!}+\frac{e^{-\lambda }{\lambda }^{n-1}}{(n-1)!}+\frac{e^{-\lambda }{\lambda }^{n-2}}{(n-2)!}+\cdots +e^{-\lambda }$

$=\sum _{i=0}^n{e}^{-\lambda }\frac{{\lambda }^i}{i!}$