improper integral convergence test does the following integral converge for all $p$?
$$\int_1^\infty e^{-x}x^p\mathrm dx$$
I believe it converges for all $p$, since exponential decays faster than polynomial. But I cannot seem to form a sequence of inequalities to prove that the integral converges.
 A: Since exponentials decay faster than powers can grow, $e^{-x/2}x^p$ is bounded by some $M>0$ on $[1,\infty)$, and therefore $$\int_1^\infty e^{-x/2}(e^{-x/2}x^p)dx\leq M\int_1^\infty e^{-x/2}dx<\infty. $$
A: 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 $p=0$. 
A: 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} $.
A: $$(p+2)!\int_1^\infty e^{-x}x^p\mathrm dx< \int_1^\infty  x^{-p-2}x^p= \int_1^\infty  x^{-2}$$, and so it converges.
Or we can use something similar to Hardy’s test for uniform convergence.  The integral can be regarded as $\int e^{-(1-\lambda)x} e^{-(\lambda)x}x^p$, where $0<\lambda<1,\ e^{-(1-\lambda)x} $ is positive and decreasing, $\int e^{-(\lambda)x}x^p< \frac{\lambda^{p+2}}{(p+2!}\int x^{-p-2}x^p $ is bounded, and so the series converges.
Either way one wants to reduce it to an integral of $x^\mu$ where $\mu<-1$. Similar to the discussion of $\sum n^{-k}$.
