Question

Question #6e061

Answer 1

`sum_(n=0)^oo1/(n!) = e`

Explanation:

Consider the development on MacLaurin series of `e^x`:

`e^x = sum_(n=0)^oo x^n/(n!)`

This series is convergent for every `x in RR`.
In particular for `x=1` we have:

`sum_(n=0)^oo1/(n!) = e^1 = e`