# Will this series with radical converge?

I'm trying to test the following series for convergence: $$\sum_{n=2}^{\infty}\frac{n}{\sqrt{n^5-n^3}}$$

I've progressed through several tests but am having trouble developing an intuition of how to approach a problem like this. I've checked the following cases so far:

• Divergence Test ($\lim a_n = 0$, so not helpful)
• Geometric Series Test (I can't find a straightforward way to find a common ratio)
• p-Series Test (it does not appear to be a p-Series)
• Limit Comparison and Comparison Tests (I can't find another series with which to prove convergence or divergence)
• Integral Test (I'm unable to integrate the expression)

Obviously I'm missing something here, but I'm just not sure which it is.

• Terms are roughly on the order of $\frac{n}{n^{5/2}} = \frac{1}{n^{3/2}}$ which should converge. How you get there is combination of comparison and $p$-series tests. Jan 23, 2018 at 14:28
• OK. It appears that I can just use the comparison test with something like $\frac{1}{x^\left(1.1\right)}$. Thank you! Jan 23, 2018 at 14:31
• Oftentimes, limit comparison is easier to use than comparison because you don't have to care if the "right" series dominates. Jan 23, 2018 at 14:33
• You should try to work it all out and answer your own question here. Would be good for you. Jan 23, 2018 at 14:33
• @Randall The trouble I have with limit comparison is that I have a hard time ending up with a constant. Seems I'm always getting infinity or infinity over infinity. Adding in the radical in the denominator makes it even easier to get lost down dead-ends. Jan 23, 2018 at 14:46

First of all, you seem to have misused some of the tests. For example, the divergence test only says that the series $\sum a_n$ diverges if $\lim a_n\neq 0$, which is not true here since $$\lim_{n\to\infty}\frac{n}{\sqrt{n^5-n^3}} = 0.$$

Second of all, you can make a comparison test:

$$\frac{n}{\sqrt{n^5-n^3}} = \frac{n}{\sqrt{n^5}\sqrt{1 - \frac{1}{n^2}}}=\frac{1}{\sqrt{n^3}}\cdot \frac{1}{\sqrt{1-\frac{1}{n^2}}}\leq 2\cdot \frac{1}{\sqrt{n^3}}$$

and of course, $$\sum_{n=1}^\infty \frac{2}{\sqrt{n^3}}$$ converges as it is two times a $p$-series.

• Thanks for the explanation and for pointing out my error with the divergence test. I'd just been checking if the sequence increases, but now I I see what I was doing wrong there. As for implementing the comparison test, I think I just need more practice so that I can "see" what you're seeing without having to try a dozen other strategies first. I appreciate your help. Jan 23, 2018 at 14:52
• @AlexJohnson In general, you just need to find the biggest factor of the numerator and denominator. In this case, the numerator is $n$, and the denominator, for large $n$, is dominated by $\sqrt{n^5}$. Since $\frac{5}{2} - 1> 1$, you have convergence.
– 5xum
Jan 23, 2018 at 14:54

You forgot the equivalence test: $$\sqrt{n^5-n^3\strut}\sim_\infty n^{5/2},\enspace\text{hence }\;\frac n{\sqrt{n^5-n^3\strut}}\sim_\infty \frac n{n^{5/2}}=\frac 1{n^{3/2}},$$ which is a convergent Riemann series.

Let $a_n$ be the $n$th term of your series and let $b_n= \frac{1}{n^{3/2}}$. Why is this my choice of $b_n$? See my first comment under your question.

Now, limit compare by $\lim_{n \to \infty} \frac{b_n}{a_n}$. Do the algebra and eventually get $$\frac{b_n}{a_n}= \frac{\sqrt{n^5-n^3}}{n^{5/2}} = \frac{\sqrt{n^5-n^3}}{\sqrt{n^5}} = \sqrt{\frac{n^5-n^3}{n^5}} = \sqrt{1-\frac{1}{n^2}}.$$ Now take the limit, and notice that it's easily $1$. Limit comparison now tells you that your series converges because $\sum_n \frac{1}{n^{3/2}}$ is a convergent $p$-series.

• I fudged a step or two but was able to get to the same place as you. This was all a tremendous help. Thank you again. Jan 23, 2018 at 15:00
• Excellent. You're well on your way. Jan 23, 2018 at 15:00

Simply note that

$$\frac{n}{\sqrt{n^5-n^3}}\sim \frac{1}{n^\frac32}$$

thus $\sum_{n=2}^{\infty}\frac{n}{\sqrt{n^5-n^3}}$ converges by comparison with

$$\sum_{n=2}^{\infty}\frac{1}{n^\frac32}$$