# How to test whether this series converge or not $\sum_{n=2}^{∞} \frac{1}{\left(n^5-n\right)^{\frac{1}{4}}}$?

How to test whether this series converge or not $$\sum_{n=2}^{∞} \frac{1}{\left(n^5-n\right)^{\frac{1}{4}}}$$

I tired using the ratio test and that didn't work, because $$\lim _{n\to \infty }\left(\frac{a_{n+1}}{a_n}\right) = 1$$ which is indeterminate by the ratio test. So I also tried using the comparison test $$0< a_n < b_n$$ and I couldn't find a suitable $$b_n$$ that I am familiar with. Or do I even have to use this? Can I just use this theorem: If a series $$\sum_{n=1}^{\infty}a_n$$ of real numbers converges then $$\lim_{n \to \infty}a_n = 0$$? When do you even use the comparison? How do you tell?

Many thanks everyone.

$$n^{5}-n\geq n^{5}-(1/2)n^{5}=(1/2)n^{5}$$, so $$\sum\dfrac{1}{(n^{5}-n)^{1/4}}\leq\sum\dfrac{2}{n^{5/4}}<\infty$$.

Note that for $$n\ge2$$, $$n^5-n>\frac12n^5$$ Hence $$\sum_{n=2}^\infty\frac1{(n^5-n)^{1/4}}<\sum_{n=2}^\infty\frac1{(n^5/2)^{1/4}}=2^{1/4}\sum_{n=2}^\infty n^{-5/4}<\infty$$

• Hi Parcly thanks for the answer. So can I say $\lim _{n\to \infty }\left(\frac{1}{\frac{1}{2}n^{\frac{5}{4}}}\right)$ = 0 which by the theorem which I included in the description, this sequence $b_n$ converges and since this is bigger than $a_n$ (our original function), $a_n$ converges too May 4 at 5:50
• Also Do I have to prove that $n^5 -n> \frac{1}{2}n^5$? May 4 at 5:51

$$\frac{1}{\left(n^5-n\right)^{\frac{1}{4}}}=\frac{1}{n^{\frac54}\left(1-\frac1{n^4}\right)^{\frac{1}{4}}}=:a_n$$, say
Note that $$\frac{a_n}{\frac {1}{n^{\frac54}}}=\frac 1{(1-\frac 1{n^4})^{\frac 14}}\to 1$$

So for big enough $$n$$, we must have $$|\frac{a_n}{\frac {1}{n^{\frac54}}}-1|\lt \frac 12\implies |a_n|\lt \frac 32\frac{1}{n^{\frac 54}}$$.

$$\sum_{n=2}^\infty\frac{1}{n^{\frac 54}}$$ converges and hence by comparison test $$\sum_{n=2}^\infty a_n$$ converges.

You can get to the answer by a combination of ratio test and comparison. Note that for sufficiently large $$n$$ we have $$n^5-n> n^{4.5},$$ hence $$\sum\frac{1}{(n^5-n)^{0.25}}<\sum\frac{1}{n^{1.125}}$$ and note that $$\sum\frac{1}{n^p}$$ converges for $$p>1$$ and diverges elsewhere.

P.S.

Investigating the behavior of a series for large $$n$$ would give you a huge idea. In this case, $$n^5$$ dominates $$n$$ and the term $$a_n$$ is somewhat $$O(n^\frac{5}{4})$$, so the series must converge. The rest of the process is to prove your intuition formally.