# convergence of a series using the root test

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.

• The third line is wrong. – Claude Leibovici Oct 30 '20 at 14:10
• @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... – Raffaele Oct 30 '20 at 14:16
• $\arccos\frac{1}{n^2}\to 1$ – Raffaele Oct 30 '20 at 14:17
• I still dont understand fully how is arcos 1/n^2 equal to one if we take the limit – GregoryStory16 Oct 30 '20 at 14:22
• @GregoryStory16. $\frac \pi 2$ – Claude Leibovici Oct 30 '20 at 14:26

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 $$1.7\times 10^{10}$$.

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 $$a_n \not \to 0$$.