Algebra

• Mar 16th 2013, 10:05 PM
qwertyuiop
Algebra

Show that for $\lambda>0$

$\sum_{i=1}^{\infty }i \times \frac{e^{-\lambda}\lambda^{i}}{i!} = \lambda$

where

$x!=\prod_{j=0}^{x=1}(x-j)=x \times (x-1)\times ... \times 1, x=1,2,3....$
$and$

$x! = 1 , x=0$
• Mar 16th 2013, 10:19 PM
Re: Algebra
You can re-write the expression as

$\sum_{i=1}^{\infty }\frac{e^{-\lambda}\lambda^{i}}{(i-1)!}$ by factorial property
$\lambda \sum_{i=1}^{\infty }\frac{e^{-\lambda}\lambda^{i-1}}{(i-1)!}$ factor out lambda not dependent on i
$\lambda \sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!}$ Let $j=i-1$ change of index.

If you can show
$\sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!}$ is equal to 1, then you are done. This sum looks extremely familiar to me, might be from statistics. I'll help you find a solution for this part if you need it online.

EDIT://

Note that $e^{-\lambda}$ is constant and that the remaining is the Taylor Series expression of $e^{\lambda}$
• Mar 16th 2013, 11:23 PM
qwertyuiop
Re: Algebra
Quote:

$\sum_{j=0}^{\infty }\frac{e^{-\lambda}\lambda^{j}}{j!}$ is equal to 1, then you are done. This sum looks extremely familiar to me, might be from statistics. I'll help you find a solution for this part if you need it online.
MacstersUndead's edit is the solution. The first line comes from the fact that $\frac{i}{i!}=\frac{i}{(i)(i-1)...(1)}=\frac{\not i}{(\not{i})(i-1)...(1)}=\frac{1}{(i-1)!}$