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)

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 $$\sum_{i=0}^{\infty} (1-\phi)^{2i} \cdot \begin{pmatrix} 1 & -\phi \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \sum_{i=0}^{\infty} (1 - \phi)^{2i} = \phi.$$
