improper integral convergence test

Since exponentials decay faster than powers can grow, $e^{-x/2}x^p$ is bounded by some $M>0$ on $[1,\infty)$, and therefore $$\int_1^\infty e^{-x/2}(e^{-x/2}x^p)dx\leq M\int_1^\infty e^{-x/2}dx<\infty. $$

Use integration by parts to relate your integral to the one with $p$ replaced with $p-1$. Repeat until the exponent is negative, then compare to the case $p=0$.

We can also get the result as follows.

Let $n$ be a positive integer such that $n-p > 1$. For any $x\geq 1$, it follows from $e^x > \frac{{x^n }}{{n!}}$ (recall that $e^x = \sum\nolimits_{n = 0}^\infty {\frac{{x^n }}{{n!}}}$) that $$ e^{ - x} x^p < \frac{{n!}}{{x^{n - p} }}. $$ Since $\int_1^\infty {\frac{n!}{{x^{n - p} }}\,dx}$ converges, so does $\int_1^\infty {e^{ - x} x^p \,dx} $.