# Thread: tricky question, prove for all positive integers n

1. ## tricky question, prove for all positive integers n

Prove that for all x > 0 and all positive integers n

$\displaystyle e^x > 1 + x + x^2/2! + x^3/3!+ ... + x^n/n!$

Hint:
$\displaystyle e^x$= 1 + $\displaystyle \int _0\,^x\!e^tdt$ > 1 + $\displaystyle \int _0\,^x\!dt$ = 1+ x

$\displaystyle e^x$= 1 + $\displaystyle \int _0\,^x\!e^tdt$ > 1 + $\displaystyle \int _0\,^x\!(1+t)dt$ = $\displaystyle 1+ x + x^2/2$, and so on.

Prove by induction on n >= 1, must jutify the hint

2. Originally Posted by wopashui
Prove that for all x > 0 and all positive integers n

$\displaystyle e^x > 1 + x + x^2/2! + x^3/3!+ ... + x^n/n!$

Hint:
$\displaystyle e^x$= 1 + $\displaystyle \int _0\,^x\!e^tdt$ > 1 + $\displaystyle \int _0\,^x\!dt$ = 1+ x

$\displaystyle e^x$= 1 + $\displaystyle \int _0\,^x\!e^tdt$ > 1 + $\displaystyle \int _0\,^x\!(1+t)dt$ = $\displaystyle 1+ x + x^2/2$, and so on.

Prove by induction on n >= 1, must jutify the hint
Excause me,
$\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \;\ \forall \;\ x \in R$
Its the well-known mac. series for $\displaystyle e^x$ !

3. Originally Posted by General
Excause me,
$\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} \;\ \forall \;\ x \in R$
Its the well-known mac. series for $\displaystyle e^x$ !
Exactly! He's simply claiming that $\displaystyle e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!} > \sum_{k=0}^{n} \frac{x^k}{k!}$...

4. Originally Posted by Defunkt
Exactly! He's simply claiming that $\displaystyle e^x=\sum_{k=0}^{\infty} \frac{x^k}{k!} > \sum_{k=0}^{n} \frac{x^k}{k!}$...

Sorry, I don't understand how you guys got that from? explain more please!