How to prove this function is increasing $\frac{e^x - x -1}{x^2}$ How to prove that the function $\displaystyle h(x) = \frac{e^x - x - 1}{x^2} = \sum_{k=2}^{\infty} \frac{x^{k-2}}{k!}$ is increasing?
Edit for context: I was asking because this function appears in Wainwright's high dimensional statistics textbook, as part of proving a Bernstein type probability inequality for random variables that are bounded on one side.
The function should be extended by continuity at $0$ with $h(0)=1/2$ and it becomes differentiable at $0$ with $h'(0)=1/6$.
For $x\ne0$, it holds
$$
h'(x)=\dfrac{(x-2)e^x+x+2}{x^3}
$$
and we can consider $g(x)=(x-2)e^x+x+2$. Then $g'(x)=(x-1)e^x+1$ and $g''(x)=xe^x$. Thus we infer that $g'$ has a minimum at $x=0$, with $g'(0)=0$. Therefore we have $g'(x)&gt;0$ for $x\ne0$, which proves that $g$ is strictly increasing.
Since $g(0)=0$, we conclude that $g(x)&lt;0$ for $x&lt;0$ and $g(x)&gt;0$ for $x&gt;0$.
Therefore $h'(x)&gt;0$ for every $x$.
For $x\not =0$ we have :
$$\int_0^1 v\exp(uvx)du=\frac{\exp(vx)-1}{x}$$ and
$$\int_0^1 \frac{\exp(vx)-1}{x}dv =\frac{1}{x}(\frac{\exp(vx)}{x}-v)_0^1=\frac{\exp(x)-1-x}{x^2}$$
Hence with $D=[0,1]\times [0,1]$
$$\int \int_Dv\exp(uvx)dudv= \frac{\exp(x)-1-x}{x^2}$$ and this is also true for $x=0$. Now use that the function $x\to v\exp(uvx)$ for fixed $u,v\in [0,1]$is increasing.