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Determine (with justification) whether the series:

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

Since this will be positive at all times I thought it was a good candiate for the Limit Comparison Test. So I let the origional function be my $a_n$, and I chose my $b_n$ as $\frac 1{n^2}$. And I know that will go towards 0 as n goes to infinity. But would that work to conclude anything about the origional series?

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  • $\begingroup$ I would recommend Direct Comparison with the same series. $\endgroup$ – Hayden Nov 18 '14 at 23:13
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hint: $$ \sqrt{n^2 - 1} = \sqrt{(n+1)(n-1)} \ge n-1 $$ so: $$ \frac 1{n\sqrt{n^2 - 1} }\le \frac 1{n(n-1)} = \frac 1{n-1} - \frac 1n $$ and $$ \sum_{k=2}^N \frac 1{n-1} - \frac 1n = 1 - \frac1N\to 1 $$is convergent.

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  • $\begingroup$ So I should just use the Comparison test? Then I could just say something like: We consider the series $a_n$. We will compare it to 1/(n(n-1)), which we know converges. And since $b_n$ is greater than $a_n$, it must also converge. Does that sound like good way to prove it? $\endgroup$ – Overclock Nov 18 '14 at 23:32
  • $\begingroup$ yes, just replace greater by its opposite :) $\endgroup$ – mookid Nov 18 '14 at 23:39
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Hint: Use the limit comparison test, with $\sum 1/n^2$ to compare against like you suspected, which gives a convergent series (which can be shown by the Cauchy condensation test). Then all you have to show is that $\lim_{n \to \infty} n/\sqrt{n^2+1} = 1$.

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We have that, for $n\ge 2$, $$ \sqrt{n^2-1}>n-1, $$ as $n^2-1>n^2-2n+1$ or $2n>2$.

Thus $$ 0<\frac{1}{n\sqrt{n^2-1}}<\frac{1}{n(n-1)}<\frac{1}{(n-1)^2}, $$ and hence $$ \sum_{n=2}^\infty\frac{1}{n\sqrt{n^2-1}}\le\sum_{n=2}^\infty\frac{1}{(n-1)^2}=\sum_{n=1}^\infty\frac{1}{n^2}<\infty. $$

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