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.

What you tried concerns ''0''. The
provided hint concerns what about $+\infty .$ Note that the exponential
function is above its tangent at $x=0,$ that is, 
$$
e^{x}\geq x+1,\ \ \ \ \ \ for\ all\ x\in 
%TCIMACRO{\U{211d} }
%BeginExpansion
\mathbb{R}
%EndExpansion
$$
then, for $t=\frac{1}{x}>0$ one has
\begin{eqnarray*}
e^{1/t} &\geq &\frac{1}{t}+1>0 \\
e^{1/t}-1 &\geq &\frac{1}{t}>0 \\
t &\geq &\frac{1}{e^{1/t}-1}>0 \\
\frac{1}{t^{2}} &\geq &\frac{1}{t^{3}(e^{1/t}-1)}>0
\end{eqnarray*}%
so since 
$$
\int_{\ln 2}^{+\infty }\frac{1}{t^{2}}dt\text{ converges}
$$
then 
$$
\int_{\ln 2}^{+\infty }\frac{dt}{t^{3}(e^{1/t}-1)}\text{  converges
too.}
$$
