Question

Approximate `sum_(n=1)^oo(-1)^n/((3^n)(n!))` correct to four decimal places?

Answer 1

`e^(-1/3)-1`

Explanation:

Think that

`e^-x = sum_(n=0)^oo (-1)^n x^n/(n!)` so making `x = 1/3` we have

`sum_(n=0)^oo (-1)^n (1/3)^n/(n!)= e^(-1/3)` and then

`sum_(n=1)^oo (-1)^n (1/3)^n/(n!)= e^(-1/3)-1`