If $\phi=\frac12(1+\sqrt5)$ and $n=\frac11+\frac1{1+\phi}+\frac1{1+\phi+\phi^2}+\cdots$, then evaluate $\lfloor2n\rfloor+\lceil2n\rceil$ The question is

Given that $\phi=\frac{1+\sqrt{5}}{2}$. Let $$n=\frac{1}{1}+\frac{1}{1+\phi}+\frac{1}{1+\phi+\phi^2}+\frac{1}{1+\phi+\phi^2+\phi^3}+\dots$$
The value of $\lfloor2n\rfloor+\lceil2n\rceil$ is $\dots$

I tried to find a general pattern that might exist for the first few terms but was unable to find a pattern that might simplify the problem. Any help would be appreciated, thanks.
 A: We will use twice the identity $\phi^2=\phi+1$ in our computation below.
Since $1+\phi+\phi^2+\cdots+\phi^{k-1}=\frac{\phi^k-1}{\phi-1}$, we have
\begin{align*}
n&=1+\frac{\phi-1}{\phi^2-1}+\frac{\phi-1}{\phi^3-1}+\frac{\phi-1}{\phi^4-1}+\cdots\\
&>1+(\phi-1)\left(\frac1{\phi^2}+\frac1{\phi^3}+\cdots\right)=1+(\phi-1)\frac{1/\phi^2}{1-1/\phi}\\
&=1+\frac1\phi=\frac{\phi+1}{\phi}=\phi.
\end{align*}
On the other hand
\begin{align*}
n&=\frac{1}{1}+\frac{1}{1+\phi}+\frac{1}{1+\phi+\phi^2}+\frac{1}{1+\phi+\phi^2+\phi^3}+\cdots\\
&<1+\frac1{1+\phi}+\frac{1}{\phi+\phi^2}+\frac{1}{\phi+\phi^2+\phi^3}+\cdots\\
&=1+\frac1{1+\phi}+\frac{1}{\phi}\left(\frac1{1+\phi}+\frac{1}{1+\phi+\phi^2}+\cdots\right)\\
&=1+\frac1{1+\phi}+\frac{1}{\phi}(n-1),
\end{align*}
hence
$$\left(1-\frac1\phi\right)n<1+\frac1{1+\phi}-\frac{1}{\phi},$$
and thus
$$n<1+\frac{\frac1{1+\phi}}{1-\frac1\phi}=1+\frac{\phi}{\phi^2-1}=2.$$
As a result, $3<\sqrt 5+1<2n<4$. Wolfram Alpha agrees with our result.
A: $$2\sum_{k=1}^5\frac{\varphi-1}{\varphi^k-1}=\frac{1105-323\sqrt5}{110}\approx3.48$$
and
$$2\sum_{k=6}^\infty\frac1{1+\varphi+\dots+\varphi^k}<2\sum_{k=6}^\infty\frac1{\varphi^k}=\frac2{\varphi^4}\approx0.29$$
hence
$$\lfloor2n\rfloor+\lceil2n\rceil=7.$$
