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Unless I've done some calculations wrong, both tests appear to be inconclusive. I have my doubts that this is the correct outcome.

I've chosen my $\sum t_n$ to be $\sum_{n=2}^{\infty}\frac{1}{n}$

Using the Direct Comparison Test:

\begin{align} \frac{1}{n\sqrt{n^2-1}}&\geq\frac{1}{n}\\ n\sqrt{n^2-1}&\leq n \end{align} Clearly this is not true for all $n\geq 2$.


Now, using the Limit Comparison Test: \begin{align} \lim_{n\to\infty}\frac{a_n}{t_n}&=\lim_{n\to\infty}\frac{1}{n\sqrt{n^2-1}}\times\frac{n}{1}\\ &=\lim_{n\to\infty}\frac{1}{\sqrt{n^2-1}}\\ &=0 \end{align} Which is again inconclusive

So did I do something wrong here, or am I doing the comparison tests incorrectly?

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    $\begingroup$ Hint: $\frac{1}{n^2}$ might be a better choice for comparison. $\endgroup$ – Peter Košinár Jun 13 '13 at 19:33
  • $\begingroup$ The most dominant term appears to be $n$ though. $\sqrt{n^2}=n$. $\endgroup$ – agent154 Jun 13 '13 at 19:34
  • $\begingroup$ Exactly so! $\sqrt{n^2}=n$, so the series actually behaves like $\frac{1}{n.n}=\frac{1}{n^2}$. Remember, it's product of the two terms, not a sum of them. $\endgroup$ – Peter Košinár Jun 13 '13 at 19:35
  • $\begingroup$ But you also have a factor of $n$ in the denominator! $\endgroup$ – Namaste Jun 13 '13 at 19:35
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Hint: try comparing your series to $$\sum_{n = 2}^\infty \dfrac 1{n^2}$$

$$a_n = \dfrac{1}{n\sqrt{n^2 - 1}} \sim \dfrac{1}{n\sqrt{n^2}} = \dfrac 1{n\cdot n} = \dfrac{1}{n^2}= t_n$$

The limit comparison text will work very nicely here.

\begin{align} \lim_{n\to\infty}\frac{a_n}{t_n}&=\lim_{n\to\infty}\frac{1}{n\sqrt{n^2-1}}\times\frac{n^2}{1}\\ &=\lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}}\\ &=1 \end{align}

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  • $\begingroup$ Why $n^2$? I was taught to pick the most dominant term, which appears to be $n$. $\endgroup$ – agent154 Jun 13 '13 at 19:35
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    $\begingroup$ Isn't the radical being multiplied by $ \ n \ $ ? $\endgroup$ – colormegone Jun 13 '13 at 19:35
  • $\begingroup$ OK, I see it now - there was one example like this in my notes that I overlooked. $\endgroup$ – agent154 Jun 13 '13 at 19:38
  • $\begingroup$ No problem...this comparison should work nicely for you! $\endgroup$ – Namaste Jun 13 '13 at 19:39
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Let $x_n=\displaystyle\frac{1}{n\sqrt{n^2-1}}\le \frac{1}{n\sqrt{n^2-2n+1}}=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$

So we have $y_m=\displaystyle \sum_{n=2}^{m}x_n\le \sum_{n=1}^{m}\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{m}<1$

And as $\displaystyle y_m=\sum_{n=2}^{m}x_n$ is increasing and bounded so it must be converge.

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  • $\begingroup$ Beautiful. (+1) $\endgroup$ – vadim123 Jun 13 '13 at 19:53

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