**Golden Section** The golden section works.

Write $u = (\sqrt{5}+1)/2$, I don't call it $\varphi$, since that
symbol is already in the problem. So of course $u^m+u^{m+1}=u^{m+2}$.

Let $p=u^{-2} \approx 0.3819$. Define:
$$
x_1 = u^{-2}, x_2=u^{-3}, x_3=u^{-4},\dots,x_m=u^{-m-1},\dots
$$
and $y_m=x_m$. Then compute
$$
\sum_{m=1}^\infty x_m = \sum_{j=2}^\infty u^{-j} =
\frac{u^{-2}}{1-u^{-1}} = \frac{u^{-2}}{u^0-u^{-1}}=
\frac{u^{-2}}{u^{-2}} = 1.
$$
Define:
$$\begin{align*}
&\varphi_{1,1}=1, \qquad\varphi_{1,j}=\varphi_{j,1}=0, j>1.
\cr
&\varphi_{2,2} = 0,\qquad\varphi_{2,j}=\varphi_{j,2}=1, j>2.
\cr
&\varphi_{m,n} = \varphi_{m-2,n-2}, m,n>2.
\end{align*}$$
Now by induction on $n$ we will show
$\sum_{m=1}^\infty x_m\varphi_{m,n} = u^{-2} = p$
for all $n$.

For $n=1$, compute
$$
\sum_{m=1}^\infty x_m \varphi_{m,1} = x_1\cdot 1 +
\sum_{m=2}^\infty x_m \cdot 0 = u^{-2}.
$$
For $n=2$, compute
$$
\sum_{m=1}^\infty x_m \varphi_{m,2} = x_1\cdot 0 + x_2\cdot 0
+ \sum_{m=3}^\infty x_m\cdot 1 =
\sum_{j=4}^\infty u^{-j} = u^{-2}.
$$
For $n>2$ apply the inductive hypothesis:
$$\begin{align*}
\sum_{m=1}^\infty x_m\varphi_{m,n} &=
x_1\cdot 0 + x_2\cdot 1 + \sum_{m=3}^\infty x_m \varphi_{m-2,n-2}
= u^{-3}+u^{-2}\sum_{j=1}^\infty x_j \varphi_{j,n-2}
\cr &=u^{-3}+u^{-2}u^{-2}=u^{-3}+u^{-4}=u^{-2}.
\end{align*}$$