Question 453400
{{{e^x = sum(x^n/n!, n = 0, infinity) = 1+0.3 + 0.3^2/2! + 0.3^3/3! + 0.3^4/4! + 0.3^5/5!}}} +... .

Using only the first five terms of this series to approximate {{{e^0.3}}}, we get 1.3498375, which is actually correct to the 4th decimal place.