Proof w/ natural log and Riemann Sum

[Xcron]

Problem states:

(A) Use mathematical induction to prove that for x\geq0 and any positive integer n.
e^x\geq1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}

(B) Use part (A) to show that e>2.7.
(C) Use part (A) to show that
\lim_{x\rightarrow\infty} \frac{e^x}{x^k} = \infty
for any positive integer k.

I thought that I could easily show that e to the x power was greater than 1 and if I could show that it was greater than 1 plus the Riemann sum:
\sum_{i=1}^n \frac{x^n}{n!} then I would have my proof...

 

[Corneo]

Base case x = 0 then the LHS becomes e^0 = 1 and the RHS becomes 1 + 0 + ... + 0. So the inequality holds. For x = k then you have e^k \geq 1 + k + \frac {k^2}{2} + ... + \frac {k^n}{n!}. Now you do x = k+1 and you have e^k e = e^{k+1} \geq e + ek + e \frac {k^2}{2} + ... + \frac {ek^n}{n!}. Clearly e>1,\ ek> k,\ e\frac {k^2}{2} > \frac {(k+1)^2}{2} and so on. Therefore

e^{k+1} \geq 1 + (k+1) + \frac {(k+1)^2}{2} + ... + \frac {(k+1)^n}{n!}

 

[HallsofIvy]

Base case x = 0 then the LHS becomes e^0 = 1 and the RHS becomes 1 + 0 + ... + 0.

No! x is a real variable so you can't do induction on x! The "induction variable" is n. The base case is for n=0. Prove that for x>= 0, e^x>= 1 (which is easy: e^x is an increasing function.)

Now, assume that e^x>= 1+ x+ (1/2)x²+ ...+ (1/k!)x^k for some k. You need to prove that e^x>= 1+ x+ (1/2)x²+ ...+ (1/k!)x^k+ (1/(k+1)!)x^k+1.

What happens if you assume that is not true, that e^x< 1+ x+ (1/2)x²+ ...+ (1/k!)x^k+ (1/(k+1)!)x^k+1 and differentiate both sides?

 

[shmoe]

What happens if you assume that is not true, that e^x< 1+ x+ (1/2)x²+ ...+ (1/k!)x^k+ (1/(k+1)!)x^k+1 and differentiate both sides?

Not much, differentiation might not preseve the inequality (I might not see what you have in mind here?)

You can go from k to k+1 by a definite integral though (I'll leave it to Xcron to set up).