Can the Limit Comparison Test Determine the Convergence of this Series?

[miglo]

Homework Statement

$$\sum_{n=2}^{\infty}\frac{1}{n\sqrt{n^2-1}}$$

Homework Equations

direct comparison test
limit comparison test

The Attempt at a Solution

so i kind of cheated and looked at the back of my book and it says to compare with $\frac{1}{n^{3/2}}$
so i tried using the direct comparison test and tried to show that the original series converges if $$\frac{1}{n\sqrt{n^2-1}}<\frac{1}{n^{3/2}}$$ since $$\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$$ is a convergent p-series test
i just don't know how to actually show $$\frac{1}{n\sqrt{n^2-1}}<\frac{1}{n^{3/2}}$$
or am i using the wrong test? limit comparison? by the way the only tests I've covered in my class are the divergence, p-series, integral, direct comparison, limit comparison tests and geometric and telescoping series

 

[susskind_leon]

you know that $n^{3/2} =n\sqrt{n}$, right?
now
$$n \sqrt{n} < n \sqrt{n^2-1}$$
$$\sqrt{n} < \sqrt{n^2-1}$$
$$n < n^2-1$$

which is valid for all n >= 2

 

[LCKurtz]

Since $n\sqrt{n^2-1}$ is of order $n^2$ this suggests the very easy limit comparison test with $\sum\frac 1 {n^2}$.