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
<br /> n \sqrt{n} &lt; n \sqrt{n^2-1}
<br /> \sqrt{n} &lt; \sqrt{n^2-1}
<br /> n &lt; 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}.