Where is this converging to on $y=x$? Well I was playing with graphs and I started plotting equations as the following:

$$\underbrace{x+y}_{degree=1}=1 \tag{1}$$
$$\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{2}$$
$$\underbrace{x^3+y^3+x^2y+xy^2}_{degree=3}+\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{3}$$
$$\underbrace{x^4+x^3y+x^2y^2+xy^3+y^4}_{degree=4}+\underbrace{x^3+y^3+x^2y+xy^2}_{degree=3}+\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{4}$$

and so on ...

And here are the plots : (Click on them to get a better pic)
Zoom:1

Zoom:2

Zoom:3


It seems like these graphs are converging to some value on the (red dashed line) $y=x$.

What is this value?


My Attempt:
Since the graphs are converging to $y=x$, hence we solve the two equations as :
(Let's take degree 2)
$$\Rightarrow x^2+y^2+xy+x+y=1 \space\space\space and \space\space\space y=x$$
$$\Rightarrow x^2+x^2+x^2+x+x=1$$
$$\Rightarrow 3x^2+2x=1$$
Hence if we take $degree=n$:

The equation becomes : $\sum\limits_{n=1}^{\infty} (n+1)x^n=1$



*

*Am I right?

*How do you solve this?

*Any more comments on this question?

*How to represent equations $(1),(2),(3),(4),\cdots$ in a more general way?

WolframAlpha showed that when solved (IDK how?) $x=1\pm \frac{1}{\sqrt{2}}$ and via plotting it looks like $x=1 - \frac{1}{\sqrt{2}}$.
How?

Thanks!!
 A: If you think of the both sides of that equation as functions of $x$, you're looking for a value $x$ with 
$$
f(x) = g(x)
$$
where $f$ is a sum and $g$ is a constant function. 
The sum happens to also be 
$$
F'(x)
$$
where 
\begin{align}
F(x) 
&= \sum_{n=1}^\infty x^{n+1}\\
&= \sum_{n=2}^\infty x^{n}\\
&= -1-x + \sum_{n=0}^\infty x^{n}\\
&= -1-x + \frac{1}{1-x}
\end{align} 
where this last step is from the geometric series formula. 
Hence 
$$
F'(x) = -1 + \frac{1}{(1-x)^2}
$$
Setting this to $1$, we get
\begin{align}
1 &=  -1 + \frac{1}{(1-x)^2}\\
2 &= \frac{1}{(1-x)^2}\\
2(1-x)^2 &= 1
(1-x)^2 &= \frac{1}{2}\\
1-x &= \pm\frac{1}{\sqrt{2}}\\
-x &= -1 + \pm\frac{1}{\sqrt{2}}\\
x &= 1 + \pm\frac{1}{\sqrt{2}}
\end{align} 
Pretty nifty!
A partial answer to question 4: 
Suppose you add a new variable, $z$, and include in your formulas enough copies of $z$ in each term to make the terms all have the same degree. So you get
\begin{align}
F_1 (x, y, z) &= x + y\\
F_2(x,y, z) &= x^2 + xy + y^2 + xz + yz \\
F_3(x, y, z) &= x^3 + x^2y + xy^2 + y^3 + x^2z + xyz + y^2 z + xz^2 + yz^2
\ldots
\end{align}
Then each $F_k$ is almost a sum of every $k$th power monomial in the three variables: 
\begin{align}
G_1 (x, y, z) &= x + y + z\\
G_2(x,y, z) &= x^2 + xy + y^2 + xz + yz + z^2 \\
G_3(x, y, z) &= x^3 + x^2y + xy^2 + y^3 + x^2z + xyz + y^2 z + xz^2 + yz^2 + z^3
\ldots
\end{align}
the only difference being the addition of the $z, z^2, z^3$ at the end of each one. 
The equations you're solving are
\begin{align}
F_1 (x, y, z) &= 1\\
F_2(x,y, z) &= 1 \\
F_3(x, y, z) &= 1\\
\ldots
\end{align}
which I propose to rewrite as 
\begin{align}
G_1 (x, y, z) &= 0 \text{ and } z = 1\\
G_2(x,y, z) &= 0 \text{ and } z = 1\\
G_3(x, y, z) &= 0 \text{ and } z = 1\\
\ldots
\end{align}
which shows that the things you've drawn are the $z = 1$ slices of certain very symmetric surfaces in 3-space, ones defined (for $k$) as the 0-level sets of the sum of all homogeneous symmetric monomials of total degree $k$. 
Is that the kind of "general" statement you were looking for? 
BTW: cool question!
