Fibonacci 数列に関する逆余接の級数
Fibonacci 数の逆数の逆余接.
前提知識 : Fibonacci 数列, 逆余接関数の加法定理, Cassini の等式
Fibonacci 数列 :
https://mathlog.info/articles/191
Cassini の等式 :
https://mathlog.info/articles/223
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Fibonacci 数列の逆数の逆余接
任意の$n\in\mathbb{Z}$に対して, $a_n=\tan^{-1}\left(\dfrac{1}{F_n}\right)$のとき$a_{2n}=a_{2n+1}+a_{2n+2}$が成りたつ.
$n=0$のときは$a_0=\pi/2$と特別に定める.
Cassini の等式から$F_{2n+1}^2-1=F_{2n+2}F_{2n}$が成立するので
$$
\begin{align}
(\mbox{右辺})&=\tan^{-1}\left(\dfrac{1}{F_{2n+2}}\right)+\tan^{-1}\left(\dfrac{1}{F_{2n+1}}\right)\\
&=\tan^{-1}\left(\frac{F_{2n+1}+F_{2n+2}}{F_{2n+1}F_{2n+2}-1}\right)\\
&=\tan^{-1}\left(\frac{F_{2n+3}}{F_{2n+1}(F_{2n+1}+F_{2n})-1}\right)\\
&=\tan^{-1}\left(\frac{F_{2n+3}}{F_{2n+2}F_{2n}+F_{2n+1}F_{2n}}\right)\\
&=\tan^{-1}\left(\frac{F_{2n+3}}{F_{2n+3}F_{2n}}\right)=a_{2n}
\end{align}
$$は恒等式である. $\quad\Box$
$\displaystyle\sum_{n=0}^\infty\tan^{-1}\left(\frac{1}{F_{2n+1}}\right)=\frac{\pi}{2}$が成りたつ.
等式$a_{2n}=a_{2n+1}+a_{2n+2}$を適用すれば
$$
\begin{align}
\sum_{n=0}^{\infty}a_{2n+1}&=\sum_{n=0}^{\infty}(a_{2n}-a_{2n+2})\\
&=a_0-\lim_{n\to\infty}a_{2n+2}=\frac{\pi}{2}
\end{align}
$$と得られる. $\quad\Box$
$\displaystyle\sum_{n=0}^\infty\tan^{-1}\left(\frac{1}{F_{n}}\right)\tan^{-1}\left(\frac{1}{F_{n+1}}\right)=\frac{\pi^2}{4}$が成りたつ.
等式$a_{2n}=a_{2n+1}+a_{2n+2}$を適用すれば
$$
\begin{align}
\sum_{n=0}^{\infty}a_{n}a_{n+1}&=\sum_{n=0}^{\infty}(a_{2n}+a_{2n+2})a_{2n+1}\\
&=\sum_{n=0}^\infty(a_{2n}+a_{2n+2})(a_{2n}-a_{2n+2})\\
&=\sum_{n=0}^\infty\left(a_{2n}^2-a_{2n+2}^2\right)\\
&=a_0^2-\lim_{n\to\infty}a_{2n+2}^2=\frac{\pi^2}{4}
\end{align}
$$と得られる. $\quad\Box$
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