Question Statement
Using Taylor's series show that $$ \lim_{t\to \infty}\int_{0}^{t} \frac{x^{2/3}}{e^{x}-1}$$ converges.
Attempt at Solution I know that from Taylor's expansions, we can expand at $x_{0}=0$ $$ f(x) = f(0) + f_{x}(0)x + \frac{f_{xx}(0)x^{2}}{2} + R_{2}(x)$$ The given function is not defined at $x_{0} =0 \ (f(0))$, so i set $x_{0} = 1$. I computed the first derivative and the second but then I get a series of constants , which I would then integrate from $0 \to t$ and get $$= \lim_{t \to \infty} \left(\frac{t}{e-1} + \frac{(e-2)t^{2}}{6(e-1)^{2}} + \frac{(1+e(2e-9))t^{3}}{18(e-1)^3} + \int_{0}^{t} R_{2}(x)\right)$$ From here I'm confused and can't seem to get further.
I also tried initially reformulating the expression as $$ \lim_{t\to \infty}\int_{0}^{t} x^{2/3} e^{-x} \left(\frac{1}{1-\frac{1}{e^{x}}}\right)$$ $$ = \sum_{k=0}^{\infty} \binom{2/3}{k} (x-1)^{k}\sum_{k=0}^{\infty} \frac{(-x)^{k}}{k!} \sum_{k=0}^{\infty} \left(\frac{1}{e^{x}}\right)^{k}$$ but with no more success.
By the Taylor series of the exponential function, $$ \frac{{x^{2/3} }}{{e^x - 1}} = \frac{{x^{2/3} }}{{x + O(x^2 )}} = \frac{1}{{x^{1/3} }} + O(x^{2/3} ), $$ which is integrable at $0$.