$$\sum_{n=1}^{\infty}\left (\sqrt{n^4+1}-n^2\right)$$

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.

  • $\begingroup$ I suggest you read this to format your maths text. $\endgroup$ – Mattos Jul 21 '15 at 2:07
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    $\begingroup$ Also $$\sqrt{n^{4} + 1} - n^{2} = \frac{1}{\sqrt{n^{4} + 1} + n^{2}} \le \frac{1}{n^{2}}$$ $\endgroup$ – Mattos Jul 21 '15 at 2:14
  • $\begingroup$ Thank you. Bookmarked link for next time. $\endgroup$ – Hq1 Jul 21 '15 at 2:14
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    $\begingroup$ @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. $\endgroup$ – Claude Leibovici Jul 21 '15 at 3:23
  • $\begingroup$ @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). $\endgroup$ – Mattos Jul 21 '15 at 3:50

\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.

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

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