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Hello I am solving the following problem.

Define a sequence of real numbers $(a_n)$ as follows: Let $a_1$ be any real number satisfying $0 < a_1 < 1$, and define $a_2, a_3,\dots$ recursively via $a_{n+1} := \cos(a_n)$ . Prove the series $\sum_n a_n$ is divergent.

Using Banachs fixed point theorem I have proven that $(a_n)$ is convergent. I am confused on how to prove the latter.

Thomas Andrews
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Lavorizia Vaughn
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1 Answers1

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All that is needed is to show that the limit of the $a_n$ is nonzero.

Thomas Andrews
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marty cohen
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  • so does this mean to write out the terms of the sequence? – Lavorizia Vaughn May 14 '21 at 00:31
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    @LavoriziaVaughn Under the assumption that $\langle a_n\rangle$ is convergent, it will have to converge to the root of the equation $\cos(x) = x ~: 0 < x < \pi/2$. The root of this equation is non-zero. – user2661923 May 14 '21 at 02:07
  • More precisely, we only need to show that $a_n$ does not converge to $0$. So we are free to assume that $a_n$ converges since otherwise that is already true. – Erick Wong Jun 02 '21 at 19:35