# 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.

• I suggest you read this to format your maths text. – Mattos Jul 21 '15 at 2:07
• Also $$\sqrt{n^{4} + 1} - n^{2} = \frac{1}{\sqrt{n^{4} + 1} + n^{2}} \le \frac{1}{n^{2}}$$ – Mattos Jul 21 '15 at 2:14
• Thank you. Bookmarked link for next time. – Hq1 Jul 21 '15 at 2:14
• @Mattos. $\sqrt{n^{4} + 1} - n^{2} = \frac{1}{\sqrt{n^{4} + 1} + n^{2}} \le \frac{1}{2n^{2}}$ could even be better for an upper bound of the sum. – Claude Leibovici Jul 21 '15 at 3:23
• @ClaudeLeibovici You're right, I just tried to give the OP an upper bound that he/she might recognise easily (even though our bounds look very similar and the OP could probably infer mine from yours). – Mattos Jul 21 '15 at 3:50

As $\sqrt{1-\dfrac{1}{n^4}}\le 1,\ \forall n\in \mathbb{N}$ we have
$$\dfrac{1}{n^2\left(\sqrt{1-\dfrac{1}{n^4}}+1\right)}\le \dfrac{1}{2n^2}$$
• You seem to have switched a "$+$" for a "$-$". Once that is fixed, your last inequality is fine. – robjohn Jul 21 '15 at 13:32
• You also need to change "As $\sqrt{1+\frac1{n^4}}\ge1$, ..." – robjohn Jul 21 '15 at 15:19
• All we really need is \begin{align} \sqrt{n^4+1}-n^2 &=\frac1{\sqrt{n^4+1}+n^2}\\ &\le\frac1{2n^2} \end{align} – robjohn Jul 21 '15 at 15:25