Convergence of a function with $e$ in the denominator $$\int^{\infty}_1\frac{dx}{x^3(e^{1/x}-1)}$$
I'm given the hint that the function $y = e^x$ has a tangent $y=x+1$ when $x=0\land y=1$.
How do I prove its convergence and find a upper-limit for the improper integral's value?
What I've tried myself
Note that $e^{1/x}-1 \ge 1 \iff e^{1/x} \ge 2$.
Hence:
$$\int^{\infty}_1\frac{dx}{x^3(e^{1/x}-1)} = \int^{1/\ln 2}_1\frac{dx}{x^3(e^{1/x}-1)} + \int_{1/\ln 2}^\infty\frac{dx}{x^3(e^{1/x}-1)}$$
Now note that our first term in RHS converges as per
$$\frac{dx}{x^3(e^{1/x}-1)} \le \frac{1}{x^3},\quad x\in[1,\ln 2]$$
And since $\int^{1/\ln 2}_1\frac{1}{x^3}$ converges, so does our first term in RHS.
Am I doing this wrong? I don't know where I'm supposed to use the hint.

Let $u=\dfrac1x$; then $x=\dfrac1u$, so $dx=-\dfrac{du}{u^2}$, and the integral becomes
$$-\int_1^0\frac{u}{e^u-1}du=\int_0^1\frac{u}{e^u-1}du\;.$$
The hint (and a look at the graph) tells you that $e^u>1+u$ for $u>0$, so $e^u-1>u$, and 
$$0<\frac{u}{e^u-1}<1$$
for $0<u\le 1$. It follows that the integral converges.
