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How to use the comparison test to prove that$$ \sum_0^\infty\arccos \left ( \frac{1}{\sqrt{1+\frac{1}{2x}}}\right)?$$was an divergent series

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  • $\begingroup$ Did you mean to start your series at $x=0$? $\endgroup$ – Simply Beautiful Art Mar 4 '17 at 16:09
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HINT:

Let $\arccos \left ( \dfrac{1}{\sqrt{1+\dfrac{1}{2x}}}\right)=y$

$\implies1+\dfrac{1}{2x}=\sec^2y\iff\cot^2y=2x\implies y=\arctan\dfrac1{\sqrt{2x}}$ as $x\not<0$

Now $$\lim_{x\to\infty}\dfrac{\arctan\dfrac1{\sqrt{2x}}}{\dfrac1{\sqrt{2x}}}=?$$

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In THIS ANSWER I showed, using standard inequalities from elementary geometry only, that the arccosine function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\sqrt{1-t^2}\le \arccos(t)\le \frac{\sqrt{1-t^2}}{t}} \tag 1$$

for $0<t\le 1$.

Using $(1)$ with $t=\sqrt{\frac{2x}{1+2x}}$ we find that

$$\arccos\left(\sqrt{\frac{2x}{1+2x}}\right)\ge \sqrt{\frac{1}{1+2x}}$$

Inasmuch as the series $\sum_{x=1}^\infty\frac{1}{\sqrt{1+2x}}$ diverges, then by comparison the series of interest diverges also.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Mar 6 '17 at 5:08
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Through Taylor expansions, we know that

$$\arccos(x)\ge\sqrt{2(1-x)}$$

Thus, we may use the comparison test:

$$\sum_{n=1}^\infty\sqrt{2\left(1-\frac1{\sqrt{1+\frac1{2n}}}\right)}$$

And by binomial expansion and limit comparison test:

$$\sqrt{1+\frac1{2n}}=1+\frac1{\sqrt{8n}}+\mathcal O(n^{-1})$$

$$1-\frac1{1+\frac1{\sqrt{8n}}+\mathcal O(n^{-1})}=\frac1{\sqrt{8n}}+\mathcal O(n^{-1})$$

$$\sqrt{\frac1{\sqrt{8n}}+\mathcal O(n^{-1})}=\frac1{\sqrt[4]{8n}}+\mathcal O(n^{-1})$$

And we can see divergence from the p-series.

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