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A series with Fibonacci

Calculus (Integrals, Series)
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Tolaso J Kos
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A series with Fibonacci

#1

Post by Tolaso J Kos » Sun Oct 16, 2016 7:18 pm

Let $F_n$ denote the $n$ - th Fibonacci number. Prove that:

$$\sum_{n=0}^{\infty} \frac{1}{F_{2n+1}+1}=\frac{\sqrt{5}}{2}$$
Hint
Don't panic! This has nothing to do with thetas. Just take a partial sum. :)
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Riemann
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Re: A series with Fibonacci

#2

Post by Riemann » Sat Oct 22, 2016 9:46 am

Hi T.

Let $\varphi=\frac{1+\sqrt{5}}{2}$ denote the golden ratio. Taking partial sum of the given series we have:

\begin{align*}
\sum_{n=0}^N\frac{1}{1+F_{2n+1}}&= \sum_{n=0}^N\frac{1}{1+\frac{\varphi^{2n+1}+\varphi^{-(2n+1)}}{\sqrt{5}}} \\
&= \sqrt{5} \sum_{n=0}^{N}\frac{\varphi^{2n+1}}{\varphi^{2(2n+1)}+\sqrt{5}\varphi^{2n+1}+1} \\
&=\sqrt{5} \sum_{n=0}^{N}\frac{\varphi^{2n+1}}{(\varphi^{2n+1}+\varphi)\left( \varphi^{2n+1}+\frac{1}{\varphi}\right)}\\
&= \sqrt{5} \sum_{n=0}^{N}\frac{\varphi^{2n+1}}{(\varphi^{2n}+1)\left( \varphi^{2n+2}+1\right)} \\
&= \frac{\varphi\sqrt{5}}{1-\varphi^2}\sum_{n=0}^N\left(\frac{\varphi^{2n}}{1+\varphi^{2n}}-\frac{\varphi^{2n+2}}{1+\varphi^{2n+2}} \right) \\
&=\sqrt{5}\left(\frac{\varphi^{2N+2}}{1+\varphi^{2N+2}} -\frac{1}{2}\right) \end{align*}

Thus:

\begin{align*}
\mathcal{S} &= \sum_{n=0}^{\infty} \frac{1}{F_{2n+1}+1}\\
&= \lim_{N \rightarrow +\infty} \sum_{n=0}^{N} \frac{1}{F_{2n+1}+1}\\
&= \lim_{N \rightarrow +\infty} \sqrt{5}\left(\frac{\varphi^{2N+2}}{1+\varphi^{2N+2}} -\frac{1}{2}\right) \\
&= \frac{\sqrt{5}}{2}
\end{align*}
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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