Stuck on a series.. Help

Apr 2014
58
0
Greece
We have the series:

\(\displaystyle \sum_{x=1}^{\infty }c \frac{l^{x}}{x!}=1\)

and we want to find the value of c.

The answer seems to be \(\displaystyle c=e^{-l}\) why is that? Is there a formula for the series?

Please help me out on this I'm confused!
 

MATHEMATICIAN

Math Team
Jul 2013
911
64
काठमाडौं, नेपाल
I think the answer is
\(\displaystyle c=\frac {1}{e^{l} -1}\)
 

skipjack

Forum Staff
Dec 2006
21,482
2,472
It's well-known that \(\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\), so \(\displaystyle \sum_{n=0}^\infty e^{-x}\frac{x^n}{n!}=1\).

Are you sure the sum in the posted problem was supposed to start at $x = 1$?
 

MATHEMATICIAN

Math Team
Jul 2013
911
64
काठमाडौं, नेपाल
It's well-known that \(\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\), so \(\displaystyle \sum_{n=0}^\infty e^{-x}\frac{x^n}{n!}=1\).
Summation starts from 1.
 
Apr 2014
58
0
Greece
Thank you both! The sum does indeed start for 0 not 1. I did a mistake there, so it all makes sense now! Thank you again.
 
Last edited by a moderator: