Question

How to find the exact value of `sum _( n=0)^(oo)2^n /(n!)`?

Answer 1

`sum_(n=0)^oo 2^n/(n!) = e^2`

Explanation:

From the MacLaurin expansion of the function `f(x) = e^x` we get that:

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

Substituting `x=2`, we have:

`e^2 = sum_(n=0)^oo 2^n/(n!)`