# Convergence for $\sum _{n=1}^{\infty }\:\frac{\sqrt[4]{n^2-1}}{\sqrt{n^4-1}}$

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?

• The polynomial $n^6+n^4-n^2-1$ is eventually increasing with an increasing derivative. Thus $\left( n^6+n^4-n^2-1 \right)^{-1/4}$ is eventually decreasing with a decreasing derivative. Wouldn't the ratio test then yield the result? Not sure about this, just a quick thought. – Matias Heikkilä Nov 11 '15 at 11:26

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.

• @AnthonyPeter Perhaps math was not aware of existing answers when preparing his/her own answer. :) – Megadeth Nov 11 '15 at 11:36
• @GudsonChou This is true. I just noticed that the times were all relatively the same, my apologies – Anthony Peter Nov 11 '15 at 11:37

Use the limit comparison test:

We have $$\frac{(n^{2}-1)^{1/4}}{(n^{4}-1)^{1/2}} = \frac{1}{(n^{2}-1)^{1/4}(n^{2}+1)^{1/2}} \sim \frac{1}{(n^{2})^{1/4}(n^{2})^{1/2}} = \frac{1}{n^{3/2}}$$ as $n \to \infty$; but the series $\sum_{n \geq 1}n^{-3/2}$ converges.

Hint: For a large enough $n$, we have $n^6 < n^6 +n^4 -n^2 -1$.

Hint: $$\frac{\sqrt[4]{n^2-1}}{\sqrt{n^4-1}}\approx \frac{\sqrt[4]{n^2}}{\sqrt{n^4}}\approx \frac{\sqrt{n}}{n^2}\approx \frac{1}{n^{\frac{3}{2}}}$$

Next, do the comparison test to the converges series $\sum \frac{1}{n^{\frac{3}{2}}}$.