Prove that the function $\frac{e^x-1-x}{x^2}$ is increasing. Prove that the function $$\frac{e^x-1-x}{x^2}$$ is increasing for all real values of $x$. 
It sounds elementary but in fact it is challenging. 
 A: On $(0,\infty)$, we may use Taylor expansion (just as noted in the comments).
We have:
$$f'(x) = \frac{e^x(x-2) + x +2}{x^3}$$
let's show that the numerator, which we'll call $g(x)$, is $\le 0$ for $x<0$. 
$$g'(x) = e^x(x-1) + 1$$
$$g''(x) = xe^x$$
Thus, $g'$ is decreasing on $(-\infty,0)$, and $\lim_{x\to -\infty}g'(x) = 1$ and $g(0) = 0$, so $g'(x) > 0$ on $(-\infty,0)$. Therefore $g$ is increasing on $(-\infty,0)$. But $\lim_{x\to -\infty} g(x) = -\infty$ and $g(0)= 0$, therefore $g(x) < 0$ on $(-\infty,0)$, and therefore $f'(x) > 0$ on $(-\infty,0)$.
A: Taking Mark's advice,
$$\frac{e^x-1-x}{x^2}=\frac12+\frac16x+\frac1{24}x^2+\dots$$
Studying monotonicity when $x>0$ is then trivial.  To tackle when $x<0$, notice that we get an alternating series with simple upper and lower bounds due to the absolute value of each term being monotone:
$$\frac12+\frac16x<\frac12+\frac16x+\frac1{24}x^2+\dots<\frac12$$
Or,
$$\sum_{n=0}^{2k-1}\frac1{(n+2)!}x^n<\sum_{n=0}^\infty\frac1{(n+2)!}x^n<\sum_{n=0}^{2k}\frac1{(n+2)!}x^n$$
From there, observe that for any $k$, these bounds are monotone as well as monotone wrt $k$ for large enough $k$, and so, by the squeeze theorem,  the function in interest is monotone when $x<0$
A: Let
$$ f(x)=\frac{e^x-1-x}{x^2} $$
and then
$$ f'(x)=\frac{e^x(x-2)+x+2}{x^2}. $$
Now we show that $f'(x)>0$ for $x<0$ or $g(x)=e^x(x-2)+x+2>0$ for $x<0$ and hence $f(x)$ is increasing for $x<0$. Note
$$ g'(x)=1+e^x(x-1)=e^x(e^{-x}+x-1)<0\text{ for }x<0$$
and hence $g(x)$ is decreasing. Therefore for $x<0$, one has $g(x)>g(0)=0$.
A: One can prove the following generalization:

For all integers $k \ge 1$ is
$$ e^x - \left(1 + x + \cdots + \frac{x^{k-1}}{(k-1)!}\right) = x^k f_k(x) $$
where the function
$$ f_k(x) = \frac{1}{(k-1)!}\int_0^1 e^{xu} (1-u)^{k-1} \, du $$
is continuous and strictly increasing on $\Bbb R$.

This follows immediately from Taylor's theorem with the integral remainder:
$$
 e^x - \left(1 + x + \cdots + \frac{x^{k-1}}{(k-1)!}\right) = \int_0^x \frac{e^t}{(k-1)!}(x-t)^{k-1} \, dt \\
 = \frac{x^{k}}{(k-1)!} \int_0^1 e^{xu} (1-u)^{k-1} \, du \, ,
$$
where we have substituted $t = xu$ in the last step.
A: Express $f(x)= \frac{e^x-1-x}{x^2}$ as a triple integral
$$f(x)=\int_0^x \int_0^t \int_0^s \frac{ue^u}{t^3}dudsdt>0
$$
