Fibonacci 数列に関する逆余接の級数

Cassini の等式から$F_{2n+1}^2-1=F_{2n+2}{F}_{2n}$が成立するので
$$\begin{array}{rl}(\text{右辺})& ={\tan }^{-1}\left({\displaystyle \frac{1}{F_{2n+2}}}\right)+{\tan }^{-1}\left({\displaystyle \frac{1}{F_{2n+1}}}\right)\\ & ={\tan }^{-1}\left(\frac{F_{2n+1}+F_{2n+2}}{F_{2n+1}{F}_{2n+2}-1}\right)\\ & ={\tan }^{-1}\left(\frac{F_{2n+3}}{F_{2n+1}(F_{2n+1}+F_{2n})-1}\right)\\ & ={\tan }^{-1}\left(\frac{F_{2n+3}}{F_{2n+2}{F}_{2n}+F_{2n+1}{F}_{2n}}\right)\\ & ={\tan }^{-1}\left(\frac{F_{2n+3}}{F_{2n+3}{F}_{2n}}\right)=a_{2n}\end{array}$$は恒等式である. $◻$

等式$a_{2n}=a_{2n+1}+a_{2n+2}$を適用すれば
$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{a}_{2n+1}& =\sum _{n=0}^{\mathrm{\infty }}(a_{2n}-a_{2n+2})\\ & =a_0-\underset{n\to \mathrm{\infty }}{lim}{a}_{2n+2}=\frac{\pi }{2}\end{array}$$と得られる. $◻$

等式$a_{2n}=a_{2n+1}+a_{2n+2}$を適用すれば
$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}{a}_n{a}_{n+1}& =\sum _{n=0}^{\mathrm{\infty }}(a_{2n}+a_{2n+2})a_{2n+1}\\ & =\sum _{n=0}^{\mathrm{\infty }}(a_{2n}+a_{2n+2})(a_{2n}-a_{2n+2})\\ & =\sum _{n=0}^{\mathrm{\infty }}(a_{2n}^2-a_{2n+2}^2)\\ & =a_0^2-\underset{n\to \mathrm{\infty }}{lim}{a}_{2n+2}^2=\frac{{\pi }^2}{4}\end{array}$$と得られる. $◻$