Originally Posted by

**Danny** Clearly it's true for $\displaystyle n = 1$. Assume true for $\displaystyle n = k$ so

$\displaystyle

e^x>1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+...+\frac {x^k}{k!}>0$ since x > 0.

Then

$\displaystyle

\int_0^x e^t \,dt> \int_0^x\left(1+t+\frac {t^2}{2!}+\frac {t^3}{3!}+...+\frac {t^k}{k!}\right)\,dt

$

This step requires, I think, full justification. It isn't true in general, of course, but it is in this case because of continuity.

Tonio

so

$\displaystyle

\left. e^t \right|_0^x > \left. t + \frac{t^2}{2!} + \frac{t^3}{3!} + \cdots + \frac{t^{k+1}}{(k+1)!}\right|_0^x

$

so

$\displaystyle

e^x -1 > x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}

$

so

$\displaystyle

e^x > 1 + x+ \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^{k+1}}{(k+1)!}

$

proving true for $\displaystyle n = k + 1$. Thus, it's true for all $\displaystyle n$.