in case if this is not to difficult I would like some help on prooving the sum of this serie.
$\displaystyle \sum_{n=0}^{fin}(-1)^n\ast {1}/{n!} = 1/e$
Re: How can I proove that series: (-1)^n (1/n!) = 1/e
Originally Posted by Melsi
in case if this is not to difficult I would like some help on prooving the sum of this serie.
$\displaystyle \sum_{n=0}^{fin}(-1)^n\ast {1}/{n!} = 1/e$
I find the question a bit strange.
Are you aware that $\displaystyle \left( {\forall x} \right)\left[ {e^x = 1 + \sum\limits_{k = 1}^\infty {\frac{{x^k }}{{k!}}} } \right]~?$
Re: How can I proove that series: (-1)^n (1/n!) = 1/e
Originally Posted by Melsi
Hello,
in case if this is not to difficult I would like some help on prooving the sum of this serie.
$\displaystyle \sum_{n=0}^{fin}(-1)^n\ast {1}/{n!} = 1/e$
Thank you in advance!
What do you know about the exponential function and "e"?