2
$\begingroup$

The other day I was at work and I came across the following question in a calculus textbook. The question is to test the following series for convergence or divergence by using the limit comparison test: $$ \sum_{n=1}^\infty {1 \over \sqrt{n^3+1}}$$ My first thought was to compare it with some type of p-series, such as: $$ \sum_{n=1}^\infty{1 \over n^{{3 \over 2}}}$$ Doing this ends up getting nowhere, because when trying to take the limit of the ratio using L'Hopital's Rule the fraction never reduces to a useful expression. After trying a few other comparison's I sought the opinion of a math professor at my college, and she was also unable to find a solution.

Does anyone have any ideas on how this series be tested using the limit comparison test?

$\endgroup$
4
$\begingroup$

If you really want to use Limit Comparison, do it with $\sum \frac{1}{n^{3/2}}$. Note that $\sqrt{n^3+1}=n^{3/2}\sqrt{1+1/n^3}$.

$\endgroup$
  • $\begingroup$ Interesting. Can you please explain how you obtained that result? Or point me somewhere that explains? $\endgroup$ – wgrenard May 15 '14 at 6:38
  • 1
    $\begingroup$ To do it the long way, $n^3+1=n^3(1+1/n^3)$. Now take the square root. $\endgroup$ – André Nicolas May 15 '14 at 6:41
  • $\begingroup$ Okay, makes sense. However, using that result the limit comparison test is inconclusive. When you take the limit of the ratio you get 0 or infinity (depending on which series you decide to put on top). $\endgroup$ – wgrenard May 15 '14 at 6:49
  • $\begingroup$ Also, I realize it is not ideal to use limit comparison. Obviously, it would be incredibly simple with direct comparison. I am curious about using limit comparison, however, simply because that was what test the student was instructed to use. $\endgroup$ – wgrenard May 15 '14 at 6:50
  • 3
    $\begingroup$ You get $1$. For $\frac{1/n^{3/2}}{1/\sqrt{n^3+1}}=\frac{\sqrt{n^3+1}}{n^{3/2}}=\frac{n^{3/2}\sqrt{1+1/n^3}}{n^{3/2}}=\sqrt{1+1/n^3}$. $\endgroup$ – André Nicolas May 15 '14 at 6:55
1
$\begingroup$

I may misunderstood where's the problem because you got your answer :
You're testing a serie of positive terms so increasing.
The last point is to see if the serie is bounded from above BUT since you've noticed that $$ {1 \over \sqrt{n^3+1}} \leq {1 \over \sqrt{n^3}} $$ then
$$ \sum_{n=1}^\infty {1 \over \sqrt{n^3+1}} \leq \sum_{n=1}^\infty {1 \over \sqrt{n^3}} $$ The RHS exists because it's a Riemann zeta function with s = 1.5 > 1. So your serie exists.

$\endgroup$
  • $\begingroup$ Just a a complement to your answer, the value of the rhs $(\zeta \left(\frac{3}{2}\right))$ is $2.61238$ and the value of the lhs is $2.29412$ $\endgroup$ – Claude Leibovici May 15 '14 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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