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I want to find whether this series diverges or converges using the root test $$ \sum_{n=1}^\infty \arccos^n \left(\frac{1}{n^2}\right) $$ $$ \sqrt[n] {\arccos^n \left(\frac{1}{n^2}\right)} $$ $$ \arccos \sqrt[n] {\left(\frac{1}{n^2}\right)} $$ $$ \arccos 1 = 0 $$ and since its 0 the series should converge but I am not sure about the calculations there.

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    $\begingroup$ The third line is wrong. $\endgroup$ – Claude Leibovici Oct 30 at 14:10
  • $\begingroup$ @ClaudeLeibovici I don't know how to contact you in a polite way. Raffaele's hypothesis is false :( so I deleted my answer. $341, 561, 645$ are counterexamples in the first thousand... $\endgroup$ – Raffaele Oct 30 at 14:16
  • $\begingroup$ $\arccos\frac{1}{n^2}\to 1$ $\endgroup$ – Raffaele Oct 30 at 14:17
  • $\begingroup$ I still dont understand fully how is arcos 1/n^2 equal to one if we take the limit $\endgroup$ – GregoryStory16 Oct 30 at 14:22
  • $\begingroup$ @GregoryStory16. $\frac \pi 2$ $\endgroup$ – Claude Leibovici Oct 30 at 14:26
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The series diverges. Indeed $$\arccos^n\left(\frac{1}{n^2}\right)\sim \left(\frac{\pi }{2}\right)^n+O\left(\frac{1}{n^2}\right);\;n\to\infty$$ The first $50$ terms sum is about $1.7\times 10^{10}$.

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Recall that

$$\arccos(x) = \frac{\pi}{2} - \arcsin(x) =\frac{\pi}{2}-x+O(x^2) \to \frac \pi 2 $$

therefore the series diverges since $a_n \not \to 0$.

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