# Stuck on a series.. Help

#### Vaki

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?

#### MATHEMATICIAN

Math Team
I think the answer is
$$\displaystyle c=\frac {1}{e^{l} -1}$$

#### skipjack

Forum Staff
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
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.

#### Vaki

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: