# Comparison tests for series

1. Oct 16, 2011

### miglo

1. The problem statement, all variables and given/known data
$$\sum_{n=2}^{\infty}\frac{1}{n\sqrt{n^2-1}}$$

2. Relevant equations
direct comparison test
limit comparison test

3. 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 dont 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

Last edited by a moderator: Oct 16, 2011
2. Oct 17, 2011

### 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

3. Oct 17, 2011

### LCKurtz

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