. This is because of Corollary 11.2.2 and
what was shown earlier that the derivative of this function is itself while e0 = 1. Thus the
question is whether this series really equals ex. You can see that this is the case by
looking at the Lagrange form of the remainder discussed earlier. In this case, it
is
eξ
-------xn+1
(n+ 1)!
where ξ is some number between 0 and x. Does this remainder term converge to 0 as
n →∞. The answer is yes because
∑∞ eξ n+1
(n-+-1)! |x|
n=1
converges by the ratio test. Indeed,
--eξ-|x|n+2 1
(n+e2ξ)!---n+1-= |x|-----→ 0
(n+1)! |x| n +2