Determine whether the series is convergent or divergent: $$\sum _{n=1}^{\infty }\:\frac{\sqrt[4]{n^2-1}}{\sqrt{n^4-1}}$$
I cannot use the integral test!
I managed to simplify my series to $\left(n^6+n^4-n^2-1\right)^{-\frac{1}{4}}$ but I'm not sure if this helps me.
Ratio test isn't working for me, I tried Raabe-Duhamel test but that would give me a really horrendous limit to solve. Wolfram says I should use comparison test, but I'm still in the dark with this one.
Any hints/thoughts?
Best Answer
Observe that $\frac{\sqrt[4]{n^2-1}}{\sqrt{n^4-1}}\approx\frac{\sqrt{n}}{n^2}\approx\frac{1}{n^{3/2}}$ for large $n$.
Setting $a_n=\frac{\sqrt[4]{n^2-1}}{\sqrt{n^4-1}}$ and $b_n=\frac{1}{n^{3/2}}$, we have $\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$. Thus, since $\sum\limits_{n=1}^{\infty}b_n<\infty$ we obtain by limit comparison test that the series $\sum\limits_{n=1}^{\infty}a_n$ is convergent.