Limit of a particular sum? From Gradshtyen and Ryzhik, 2.111(3), for $t$ real and $\alpha$ a real constant, the indefinite integral
$$f\left( {t;\alpha } \right) = \int {dt\left( {\frac{{{t^n}}}{{t + \frac{1}{\alpha }}}} \right)}  = \frac{{{{\left( { - 1} \right)}^n}}}{{{\alpha ^n}}}\ln \left( {t + \frac{1}{\alpha }} \right) + \sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{n - k}}{t^k}}}{{k\,{\alpha ^{n - k}}}}}$$
The integral implies that
$$\mathop {\lim }\limits_{\alpha  \to 0} f\left( {t;\alpha } \right) = \mathop {\lim }\limits_{\alpha  \to 0} \int {dt\frac{{\alpha {t^n}}}{{\alpha t + 1}}}  = 0$$
Is there any way of finding the equivalent limit for the sum without recognizing the integral form?
 A: Your first expression is false. What is true is that
$$f\left(t;\alpha\right) = \int dt\left( \frac{t^n}{t + \frac 1\alpha}\right)  = \frac{\left(- 1\right)^n}{\alpha ^n}\ln\left|t + \frac 1\alpha\right| + \sum\limits_{k = 1}^n \frac{\left(-1\right)^{n - k}t^k}{k\,\alpha^{n - k}} \color{red}{+ C}$$
Indefinite integrals are only defined up to an arbitrary constant. We do not know the value of $C$. The value of $C$ is not even defined by the information given.
Actually, even this is not entirely true. The integral is improper at $t = - 1/\alpha$. Indeed, $|f|$ is infinite at this point. The value of $C$ could be different for $t < -1/\alpha$ than it is for $t > -1/\alpha$, since $f$ is already discontinuous there.
And finally, $$\lim_{\alpha \to 0} \left[\frac{\left(- 1\right)^n}{\alpha ^n}\ln\left|t + \frac 1\alpha\right| + \sum\limits_{k = 1}^n \frac{\left(-1\right)^{n - k}t^k}{k\,\alpha^{n - k}}\right] = \pm\infty$$
not $0$. The sign depends on the value of $n$ and on whether $\alpha$ is approaching $0$ from above or below.
