What is $\lim\limits_{n\to\infty}\sum\limits^n_{i=0}F_{2i}-\phi F_{2i-1}$? 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)
 A: 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}
A: 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 $$\sum_{i=0}^{\infty} (1-\phi)^{2i} \cdot \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \sum_{i=0}^{\infty} (1 - \phi)^{2i} = \phi.$$
A: $\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=1+\sum^n_{i=1}F_{2i}-\phi F_{2i-1}$
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, $$\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=1-(1-\phi)=\phi$$
