TITLE: What is $\lim\limits_{n\to\infty}\sum\limits^n_{i=0}F_{2i}-\phi F_{2i-1}$?
QUESTION [3 upvotes]: 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)

REPLY [1 votes]: $\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=1+\sum^n_{i=1}F_{2i}-\phi F_{2i-1}$
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, $$\lim_{n\to\infty}\sum^n_{i=0}F_{2i}-\phi F_{2i-1}=1-(1-\phi)=\phi$$
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