Prove convergence of improper integral Please, give-me a hint to prove that 
$$\int_{1}^{+\infty} \frac{x \, dx}{1-e^x}$$ converges. I've tried to substitute $e^x$ for $y$ and integrate by parts, but it didn't work. Thank you!
 A: One approach would be to use that $\displaystyle\frac{x}{1-e^x}=-\frac{x}{e^x-1}$,
and$\;\;\displaystyle\frac{x}{e^x-1}<\frac{2x}{e^x}=2xe^{-x}$ for $x\ge1$ since $2<e^x$ for $x\ge 1$.
A: You have received good answers to your question.   
Just for your curiosity, I shall add here that the antiderivative of the integrand you consider has a closed form solution which is given by $$-\text{Li}_2\left(e^x\right)+\frac{x^2}{2}-x \log \left(1-e^x\right)$$ 
Integrated between $1$ and $\infty$, the result is then $$-\text{Li}_2\left(\frac{1}{e}\right)-1+\log (e-1)$$
A: if $f$ is montonic then you get this:
$$ \int\limits_1^\infty f(x)\text{ d}x<\infty \quad \Leftrightarrow \quad \sum_{i=1}^\infty f(i) <\infty $$ 
it's not that hard to prove and maybe it does help!
A: By Taylor's theorem $e^x-1 > x^3/6$, so
$$\int_1^\infty \frac{x}{e^x-1} dx \leq 6 \int_1^\infty \frac{1}{x^2} dx=6$$
A: For $x\ge 5$, $e^x-1 \ge x^3$ so $\dfrac{x}{|1-e^x|} \le \dfrac{1}{x^2}$
Then
$$ \left|\int_1^{\infty} \dfrac{x}{1-e^x} dx\right| \le \int_1^{\infty} \dfrac{x}{|1-e^x|} dx \le \int_1^5\frac{x}{e^x-1} dx + \int_5^{\infty} x^{-2} \, dx$$
$$ = \int_1^5\frac{x}{e^x-1} \, dx + \dfrac{1}{5}$$
A: Hint: note that, for large $x$, 

$$ \frac{x}{1-e^x} \sim \frac{x}{-e^x} = -xe^{-x}.$$

