<s>[INST] How to show that his series converges or diverges using LCT or CT? $$\sum_{n=1}^{\infty}\left (\sqrt{n^4+1}-n^2\right)$$
The question states that either the limit comparison or comparison test can be used to determine whether the series converge or diverge. I tried finding a $B_n$ in order to test $\frac{A_n}{B_n}$ for the limit comparison but having trouble coming up with $B_n$ that I know will converge or diverge. Maybe I'm going about this the wrong way. Any help would be appreciated.  
 [/INST] \begin{align}
\sqrt{n^4+1}-n^2 &=\dfrac{\sqrt{n^4+1}-n^2}{1} \\
&=\dfrac{\left(\sqrt{n^4+1}-n^2\right)\left(\sqrt{n^4+1}+n^2\right)}{\left(\sqrt{n^4+1}+n^2\right)} \\
&=\dfrac{\left(\sqrt{n^4+1}\right)^2-n^4}{\sqrt{n^4\left(1-\dfrac{1}{n^4}\right)}+n^2} \\
&=\dfrac{n^4+1-n^4}{n^2\,\sqrt{1-\dfrac{1}{n^4}}+n^2} \\
&=\dfrac{1}{n^2\left(\sqrt{1-\dfrac{1}{n^4}}+1\right)}
\end{align}
As $\sqrt{1-\dfrac{1}{n^4}}\le 1,\ \forall n\in \mathbb{N}$ we have
$$\dfrac{1}{n^2\left(\sqrt{1-\dfrac{1}{n^4}}+1\right)}\le \dfrac{1}{2n^2}$$
therefore the series converges because the greater series converges.
</s>