一个含有Fibonacci Number的级数

\[\Large\displaystyle \sum_{n=0}^\infty \frac{1}{F_{2n+1}+1}=\frac{\sqrt5}{2}\]

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\(\Large\mathbf{Proof:}\)
Let \(\phi=\dfrac{1+\sqrt{5}}{2}\) denote the golden ratio. Then consider the partial sum,
\[\begin{align*} \sum_{n=0}^N\frac{1}{1+F_{2n+1}}&= \sum_{n=0}^N\frac{1}{1+\dfrac{\phi^{2n+1}+\phi^{-(2n+1)}}{\sqrt{5}}} \\ &= \sqrt{5} \sum_{n=0}^{N}\frac{\phi^{2n+1}}{\phi^{2(2n+1)}+\sqrt{5}\phi^{2n+1}+1} \\ &=\sqrt{5} \sum_{n=0}^{N}\frac{\phi^{2n+1}}{(\phi^{2n+1}+\phi)\left( \phi^{2n+1}+\dfrac{1}{\phi}\right)}\\ &= \sqrt{5} \sum_{n=0}^{N}\frac{\phi^{2n+1}}{(\phi^{2n}+1)\left( \phi^{2n+2}+1\right)} \\ &= \frac{\phi\sqrt{5}}{1-\phi^2}\sum_{n=0}^N\left(\frac{\phi^{2n}}{1+\phi^{2n}}-\frac{\phi^{2n+2}}{1+\phi^{2n+2}} \right) \\ &=\sqrt{5}\left(\frac{\phi^{2N+2}}{1+\phi^{2N+2}} -\frac{1}{2}\right) \end{align*}\]
Let \(N\to\infty\) to get
\[\Large\boxed{\displaystyle \sum_{n=0}^\infty\frac{1}{1+F_{2n+1}}=\color{blue}{\frac{\sqrt{5}}{2}}}\]

转载于:https://www.cnblogs.com/Renascence-5/p/5458328.html