# Testing Convergence With Limit Comparison Test

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?

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

• Interesting. Can you please explain how you obtained that result? Or point me somewhere that explains? – wgrenard May 15 '14 at 6:38
• To do it the long way, $n^3+1=n^3(1+1/n^3)$. Now take the square root. – André Nicolas May 15 '14 at 6:41
• 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). – wgrenard May 15 '14 at 6:49
• 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. – wgrenard May 15 '14 at 6:50
• You get $1$. For $\frac{1/n^{3/2}}{1/\sqrt{n^3+1}}=\frac{\sqrt{n^3+1}}{n^{3/2}}=\frac{n^{3/2}\sqrt{1+1/n^3}}{n^{3/2}}=\sqrt{1+1/n^3}$. – André Nicolas May 15 '14 at 6:55

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.

• 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 $2.29412$ – Claude Leibovici May 15 '14 at 7:30