Does $\int_{3}^{\infty}\frac{1}{(x-2)^{3/2}}\text{ d}x$ converge? I am trying to see whether or not 
$$\int\limits_{3}^{\infty}\dfrac{1}{(x-2)^{3/2}}\text{ d}x$$
converges. My first instinct was to notice in $[3, \infty)$ that 
$$\dfrac{1}{(x-2)^{3/2}} > \dfrac{1}{x^{3/2}}\text{.} $$
But $\displaystyle\int\limits_{3}^{\infty}\dfrac{1}{x^{3/2}}\text{ d}x$ converges, so that does not give me any helpful information. 
As I started typing this question, I thought of the following idea: does it suffice to show that
$$\lim\limits_{t\to \infty}\int\limits_{3}^{t}\dfrac{1}{(x-2)^{3/2}}\text{ d}x$$
exists?
 A: $$\int\limits_3^\infty\frac1{(x-2)^{3/2}}dx=\left.-2(x-2)^{-1/2}\right|_3^\infty:=-2\lim_{b\to\infty}\left(\frac1{\sqrt{b-2}}-1\right)=2$$
A: Other answers are perfectly legitimate.  Here's another way to think about it:
Note that $\displaystyle \int_3^\infty \frac{1}{(x-2)^{3/2}}dx = \int_1^\infty \frac{1}{x^{3/2}}dx = \int_1^3 \frac{1}{x^{3/2}}dx + \int_3^\infty \frac{1}{x^{3/2}}dx$.
You know already that $\displaystyle \int_3^\infty \frac{1}{x^{3/2}}dx$ converges.  Therefore, you need only show that $\displaystyle \int_1^3 \frac{1}{x^{3/2}}dx$ converges.  Thinking about this as area under a curve, this is pretty easy to see (or you could just work it out).
Thus, the value of original definite integral is the sum of two finite values, so it must also be finite (convergent).
A: Yes the given integral is convergent since
$$\frac1{(x-2)^{3/2}}\sim_\infty\frac1{x^{3/2}}$$
and the integral
$$\int_3^\infty \frac{dx}{x^{3/2}}$$
is convergent.
A: The existence of $$\lim_{t\to\infty}\int_3^t\frac1{(x-2)^{3/2}}dx$$ is exactly what is meant by the statement that $$\int_3^\infty\frac1{(x-2)^{3/2}}dx$$ converges. But you don’t have to make an indirect argument, as you can evaluate the limit (i.e., the improper integral) directly: just substitute $u=x-2$.
