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.
 A: $n^{5}-n\geq n^{5}-(1/2)n^{5}=(1/2)n^{5}$, so $\sum\dfrac{1}{(n^{5}-n)^{1/4}}\leq\sum\dfrac{2}{n^{5/4}}<\infty$.
A: Note that for $n\ge2$,
$$n^5-n>\frac12n^5$$
Hence
$$\sum_{n=2}^\infty\frac1{(n^5-n)^{1/4}}<\sum_{n=2}^\infty\frac1{(n^5/2)^{1/4}}=2^{1/4}\sum_{n=2}^\infty n^{-5/4}<\infty$$
A: $\frac{1}{\left(n^5-n\right)^{\frac{1}{4}}}=\frac{1}{n^{\frac54}\left(1-\frac1{n^4}\right)^{\frac{1}{4}}}=:a_n$, 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.
A: 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.
