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.

The integral is from $1$ to $\infty$, and your argument is for the integral from $1$ to $1/\ln2$. The integral is improper onli at $\infty$. As $x\to\infty$, $e^{1/x}-1\sim 1/x$ (here is where you use the hint.) Then, as $x\to\infty$
$$
\frac{1}{x^3(e^{1/x}-1)}\sim\frac{1}{x^2}.
$$
Notation: $f(x)\sim g(x)$ means $f(x)/g(x)\to1$  as $x\to\infty$.
