Mathematical Induction

Induction on n:

When n=1, we want to prove e^x>=1+x.

So let f(x)=e^x-x-1, f'(x)=e^x-1>=0 when x>=0,
thus f(x) is strictly increasing and so f(x)>=f(0)=0.
Therefore e^x>=1+x.
The statement is true for n=1.

Assume the statement is true for n=k,
i.e. e^x>=1+x+x^2/2!+...+x^k/k! for x>=0

When k=n+1,
by assumption we have e^t>=1+t+t^2/2!+...+t^k/k! (just replace x by t)

Integrate both side w.r.t. t from 0 to x,
∫(from 0 to x) e^t dt >= ∫(from 0 to x) [1+t+t^2/2!+...+t^k/k!] dt
(As f(x)>=g(x) in an interval I implies ∫f(x)dx>=∫g(x)dx in that interval I)

therefore it becomes
e^x-1>=x+x^2/2!+...+x^k/k!+x^(k+1)/(k+1)!
i.e. e^x>=1+x+x^2/2!+...+x^k/k!+x^(k+1)/(k+1)! when x>=0

The statement is true for n=k+1.

By the principle of M.I., the statement is true for x>=0 and positive integer n.

Remark 1. The key step is that you should apply the integration from 0 to x.
If you know this step, then the flow of the whole proof becomes clear.

Remark 2. The proof beyond me, i.e. #001, is totally wrong...since in this proof, we cannot simply apply the Taylor Series, other there is nothing to prove.

e^x >= 1 + x + x^2/(2!) + .......+x^n / (n!)
For x = 1,
L.H.S. = e^1 = e = 2.718
R.H.S. = 1 + 1 = 2
Therefore L.H.S. > R.H. S.
Assume n = k is true,
e^x >= 1 + x + x^2/(2!) + ... + x^k/(k!)
For n = k+1
L.H.S. = e^x = 1 + x + x^2/2! +... + x^k/k! + x^(k+1)/(k+1)! +x^(k+2)/(k+2)! + ... +x^∞/(∞!)
>= 1 + x + x^2/2! + ... + x^k/k! + x^(k+1) = R.H.S.
When n=K+1, L.H.S. >= R.H.S. It is true for n = k+1
By M.I. , it is true for all positive integer n