1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Matias Heikkilä Nov 11 '15 at 11:26
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @AnthonyPeter Perhaps math was not aware of existing answers when preparing his/her own answer. :) $\endgroup$ – Megadeth Nov 11 '15 at 11:36
  • $\begingroup$ @GudsonChou This is true. I just noticed that the times were all relatively the same, my apologies $\endgroup$ – Anthony Peter Nov 11 '15 at 11:37
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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}}} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.