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I give a way that may work for odd numbers. This is too long for a comment. First, the quantity $$\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big) = 2\cos\Big(\pi\Big(\frac{m}{p}+\frac{n}{ …

My impression is that you have developed $(x_0-\sqrt{S})^{2^n}$ with Newton binomial formula, separated the terms with even index from the terms with odd index to get an expression of the form $A_n-\s …

I am surprised by the question. The sums are bounded if $\alpha \ne 1$, since for every $n \ge 1$, $$\Big|\sum_{k=1}^n \Re R_\alpha^k z \Big| = \Big|\Re \Big(\sum_{k=1}^n R_\alpha^k z \Big) \Big| = \B …

Partial answer. I imagine that you mean a map $\overline{G}$ which sends $([0;a_1,a_2,\ldots],[0;a_0,a_{-1},\ldots])$ on $([0;a_2,a_3,\ldots],[0;a_1,a_0,\ldots])$. If yes, we have a simple formula $$\ …

For $-1<h<1$, $$(1+h)^{3/2}+(1-h)^{3/2} = 2 \sum_{n=0}^{+\infty}{3/2 \choose 2n}h^{2n},$$ where $${3/2 \choose 2n} = \prod_{k=1}^ {2n} \frac{5/2-k}{k}.$$ For $n \ge 1$, since $(-1)^{2n-2}=1$, we get $ …

I prove the convergence of the series. For $n \ge 1$, let $$S_n = \sum_{k=1}^n k^{-i\beta-\alpha} \text{ and } S'_n = \sum_{k=1}^n (-1)^ {k-1} k^{-i\beta-\alpha}$$ Then $$S_{2n}-S'_{2n} = 2 \sum_{k=1} …