need to find the sum of (1/n!)
from n= 1 to infinity
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need to find the sum of (1/n!)
from n= 1 to infinity
$\displaystyle \sum_{n=1}^\infty \frac{1}{n!} = \sum_{n=1}^\infty \frac{1^n}{n!} = (\sum_{n=0}^\infty \frac{1^n}{n!}) - 1 = e^1 - 1 = e - 1$
thank alout