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.

$\frac{1}{\left(n^5-n\right)^{\frac{1}{4}}}=\frac{1}{n^{\frac54}\left(1-\frac1{n^4}\right)^{\frac{1}{4}}}=:a_n$, 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.
