Will this series with radical converge?

Question

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.

Answer 1

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 $$\lim_{n\to\infty}\frac{n}{\sqrt{n^5-n^3}} = 0.$$

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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

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