# study the behavour of the series at the changing of alpha [closed]

studio della serie

$$\sum_{n=2}^{\infty}\frac{\sqrt{n^4+1} - n^2}{\ln^{\alpha} (n)}$$

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I made an adjustment to your formatting. Let me know if it is wrong. Also, what is your question? What have you thought of/done? math.stackexchange.com is not a site where you can post whatever you want and expect an answer. We need your considerations and questions. –  Rustyn Jan 26 at 19:31
The language of the site is English. Questions should be a little more detailed than this. –  Beni Bogosel Jan 26 at 19:41
@Beni Hmmm... This was a subject of debate some time ago. –  Did Jan 26 at 20:12
@Did: Ok.... I missed that discussion. –  Beni Bogosel Jan 26 at 20:21

## closed as not a real question by Micah, Nate Eldredge, Brandon Carter, Norbert, ThomasJan 26 at 21:52

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Recall that $\sqrt{1+u}=1+\frac{u}{2}+O(u^2)$ as $u$ approaches $0$.

It follows that $$\sqrt{n^4+1}=n^2\sqrt{1+1/n^4}=n^2\left( 1+\frac{1}{2n^4}+ O\left(\frac{1}{n^8}\right) \right)=n^2+\frac{1}{2n^2}+ O\left(\frac{1}{n^6}\right).$$

Therefore the general term of your series verifies $$\frac{\sqrt{n^4+1}-n^2}{(\ln n)^\alpha}=\frac{1}{2n^2(\ln n)^\alpha}+ O\left(\frac{1}{n^6 (\ln n)^\alpha} \right).$$

Now use the fact that the series $$\sum_{n\geq 1} \frac{1}{n^p(\ln n)^\alpha}$$ converges for all $p>1$ to conclude that your series converges, no matter what the value of $\alpha$ is.

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Hint: Use the identity $\sqrt{n^4+1}-n^2=\dfrac1{\sqrt{n^4+1}+n^2}$ and the equivalent $\sqrt{n^4+1}\sim n^2$ to deduce that the $n$th term of the series is equivalent to $\dfrac1{2n^\color{red}{2}\log^\alpha n}$. Since $\color{red}{2}\gt1$, the series converges for every $\alpha$.