Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Hint: $\frac{1}{n^2}$ might be a better choice for comparison. – Peter Košinár Jun 13 '13 at 19:33
The most dominant term appears to be $n$ though. $\sqrt{n^2}=n$. – agent154 Jun 13 '13 at 19:34
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. – Peter Košinár Jun 13 '13 at 19:35
But you also have a factor of $n$ in the denominator! – amWhy Jun 13 '13 at 19:35
up vote 4 down vote accepted

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}

share|cite|improve this answer
Why $n^2$? I was taught to pick the most dominant term, which appears to be $n$. – agent154 Jun 13 '13 at 19:35
Isn't the radical being multiplied by $ \ n \ $ ? – RecklessReckoner Jun 13 '13 at 19:35
OK, I see it now - there was one example like this in my notes that I overlooked. – agent154 Jun 13 '13 at 19:38
No problem...this comparison should work nicely for you! – amWhy Jun 13 '13 at 19:39

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.

share|cite|improve this answer
Beautiful. (+1) – vadim123 Jun 13 '13 at 19:53
Thanks a lot @vadim123 – Abhra Abir Kundu Jun 13 '13 at 19:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.