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Firstly, the question is:

Evaluate

$$(1-0\phi)+(2-1\phi)+(5-3\phi)+(8-5\phi)+\cdots=\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}$$

where $F_0=1,$ $F_1=1$ and $F_n=F_{n-1}+F_{n-2}$ for all integer n (even n negative).

I attempted the basic manipulations with Fibonacci numbers but ultimately I reached this:

$$\begin{align} &\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}\\ =&\lim_{n\to\infty}\sum^n_{i=0}F_{2i-1}+F_{2i-2}-\phi F_{2i-2}-\phi F_{2i-3}\\ =&\lim_{n\to\infty}1+\sum^n_{i=0}(F_{i}-\phi F_{i})+F_{n+1}\\ =&\lim_{n\to\infty}1+\sum^n_{i=0}(\frac{1-\sqrt{5}}{2}F_{i})+F_{n+1}\\ \end{align} $$

which probably doesn't get me anywhere.

(I'd also like the sum

$$\lim_{n\to\infty}\sum^n_{i=0}F_{2i-1}-\phi F_{2i-2}$$

but if the method works for this then there's no reason to explicitly calculate this for me)

J. M. ain't a mathematician
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Wen
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  • Aren't those $F_{2n}$'s and the like actually supposed to be $F_{2i}$'s etc.? – giobrach Sep 05 '17 at 01:30
  • Oops, yeah. I was sure I had checked them... – Wen Sep 05 '17 at 01:31
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    Your $F_n$ is the standard fibonacci number $f_{n+1}$ and there's the Binet's formula $f_n = \frac{\alpha^n - \beta^n}{\sqrt{5}}$ where $\alpha = \phi, \beta = -\frac{1}{\phi}$. – achille hui Sep 05 '17 at 01:49

3 Answers3

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Using $$F_n=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\sqrt5}$$ \begin{align} &1+\sum_{n=1}^{\infty} (F_{2n}-\phi F_{2n-1})\\ =&1+\sum_{n=1}^{\infty} \left(\frac{\phi^{2n+1}-(-\phi)^{-2n-1}}{\sqrt5}-\phi\frac{\phi^{2n}-(-\phi)^{-2n}}{\sqrt5}\right)\\ =&1+\sum_{n=1}^{\infty} \left(\frac{\phi^{2n+1}-(-\phi)^{-2n-1}}{\sqrt5}-\frac{\phi^{2n+1}+(-\phi)^{-2n+1}}{\sqrt5}\right)\\ =&1-\sum_{n=1}^{\infty}(1+\phi^2)\frac{(-\phi)^{-2n-1}}{\sqrt5}\\ =&1+\frac{1+\phi^2}{\sqrt5\phi}\sum_{n=1}^{\infty}\left(\frac1{\phi^2}\right)^n\\ =&1+\frac{1+\left(\frac1{\phi^2}\right)}{\sqrt5\phi}\sum_{n=0}^{\infty}\left(\frac1{\phi^2}\right)^n\\ =&1+\frac{1}{\sqrt5\phi}\frac{1+\left(\frac1{\phi^2}\right)}{1-\left(\frac1{\phi^2}\right)}\\ =&1+\frac{1}{\sqrt5\phi}\frac{\phi^2+1}{\phi^2-1}\\ =&\phi \end{align}

Kay K.
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  • The $F_n$ in question are not standard Fibonacci numbers (like what you are using), it starts with $F_0, F_1, F_2 = 1,1,2$, you need to adjust your result for that. – achille hui Sep 05 '17 at 02:12
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    For some reason I thought that this formula would give a big mess. Thanks for your help (I got the general gist of this, even though the algebra was a bit off) – Wen Sep 05 '17 at 02:22
  • @achillehui Thanks. I corrected accordingly. – Kay K. Sep 05 '17 at 03:17
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Since $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_i \\ F_{i-1} \end{pmatrix} = \begin{pmatrix} F_{i+1} \\ F_i \end{pmatrix},$$ you are looking for the limit $$\lim_{N \rightarrow \infty} \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} F_{2i} \\ F_{2i-1} \end{pmatrix} = \sum_{i=0}^N \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{2i} \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ Since $$\begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1-\phi & 1 \end{pmatrix} = (1-\phi) \cdot \begin{pmatrix} 1 & -\phi \end{pmatrix},$$ this is the series $$\sum_{i=0}^{\infty} (1-\phi)^{2i} \cdot \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \sum_{i=0}^{\infty} (1 - \phi)^{2i} = \phi.$$

user16394
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$\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=1+\sum^n_{i=1}F_{2i}-\phi F_{2i-1}$

For $i\geq 1$,$$F_{2i}-\phi F_{2i-1}=F_{2i+1}-F_{2i-1}-\phi F_{2i}+\phi F_{2i-2} \\=(F_{2i+1}-\phi F_{2i})-(F_{2i-1}-\phi F_{2i-2})$$

Then, $$\sum^n_{i=1}F_{2i}-\phi F_{2i-1}=\sum_{i=1}^{n}[(F_{2i+1}-\phi F_{2i})-(F_{2i-1}-\phi F_{2i-2})]\\=(F_{2n+1}-\phi F_{2n})-(F_1-\phi F_0)$$

$$\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=\lim_{n\to\infty}(F_{2n+1}-\phi F_{2n})-(1-\phi)$$

$F_{2n+1}-\phi F_{2n}=\dfrac{\phi^{2n+2}-\phi^{-(2n+2)}}{\sqrt5}-\phi\dfrac{\phi^{2n+1}+\phi^{-(2n+1)}}{\sqrt5}=-\dfrac{(1+\phi^2)\phi^{-(2n+2)}}{\sqrt5}$

Then $\lim_{n\to\infty}F_{2n+1}-\phi F_{2n}=0$ (since $\dfrac{1}{\phi^2}<1$)

Hence, $$\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=1-(1-\phi)=\phi$$

MAN-MADE
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