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.
 A: Set $t=1/x$. We then obtain $dt=-dx/x^2$. Hence, the integral becomes
$$I = \int_1^0 \dfrac{t}{e^t-1} (-dt) = \int_0^1 \dfrac{t}{e^t-1} dt$$
The convergence is now obvious, since for $t\in(0,1)$, we have $e^t-1 > t$, which implies
$$\dfrac{t}{e^t-1} < 1 \implies I < \int_0^1 1 dt = 1$$

We can in fact evaluate this integral
$$I = \int_0^1\dfrac{te^{-t}}{1-e^{-t}}dt = \int_0^1 \sum_{k=0}^{\infty}te^{-(k+1)t}dt = \sum_{k=1}^{\infty} \int_0^1 te^{-kt}dt$$
We have
$$\int_0^1te^{-kt}dt = \dfrac{1-e^{-k}(k+1)}{k^2}$$
Hence, the integral is
$$\sum_{k=1}^{\infty} \dfrac{1-e^{-k}(k+1)}{k^2} = \sum_{k=1}^{\infty} \dfrac1{k^2} - \sum_{k=1}^{\infty} \dfrac{e^{-k}}{k} - \sum_{k=1}^{\infty} \dfrac{e^{-k}}{k^2}=\zeta(2) + \log(1-1/e) - \text{Li}_2(1/e)$$
A: 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 
$$
\int_{\ln 2}^{+\infty }\frac{dt}{t^{3}(e^{1/t}-1)}\text{  converges
too.}
$$
A: 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.
A: 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 $x\to\infty$.
