<s>[INST] What value does the area of the region enclosed by $\sum_{k=1}^m (x^k+y^k)=2$ converge to as $m\rightarrow\infty$? Find the area enclosed by the curve
$$\sum_{k=1}^m (x^k+y^k)=2$$
where $m$ is even and $m \rightarrow\infty$.
That is for a sufficiently large even number $m$, we have:
$$x^1+y^1+x^2+y^2+x^3+y^3+\dots+x^m+y^m=2$$
See the graph of the curve ($m=250$): the $250$ here is to illustrate the region, but the required area to be calculated when $m$ is even that is tending to infinity.


WHAT I DID:
I tried to convert the equation of the curve, which has an infinite series, into a function of $x$ where $x>0 ,y>0$.
After very long and difficult steps, I got the area $(A)$ enclosed by that curve as:
$$A=\int_0^{\frac{12}{7}}\frac{7x-12}{4x-7}dx=3-\frac{1}{8}\ln(7)\approx 2.75676 \text{ unit}^2$$.

I. Is my answer right?
II. How can we solve it easily? I used $5$ pages to solve it.

Any help would be really appreciated. THANKS!
 [/INST] Doing some obvious stuff.
$\begin{array}\\
2
&=\sum_{k=1}^m (x^k+y^k)\\
&=\sum_{k=1}^m x^k+\sum_{k=1}^m y^k\\
&=\dfrac{x-x^{m+1}}{x-1}+\dfrac{y-y^{m+1}}{y-1}\\
&\to \dfrac{x}{x-1}+\dfrac{y}{y-1}
\qquad\text{as }m \to \infty\\
&= \dfrac{x-1+1}{x-1}+\dfrac{y-1+1}{y-1}\\
&= 1+\dfrac{1}{x-1}+1+\dfrac{1}{y-1}\\
\text{so}\\
0
&= \dfrac{1}{x-1}+\dfrac{1}{y-1}\\
&= \dfrac{(y-1)+(x-1)}{(x-1)(y-1)}\\
&= \dfrac{x+y-2}{(x-1)(y-1)}\\
&\text{just a line}\\
2
&=\dfrac{x-x^{m+1}}{x-1}+\dfrac{y-y^{m+1}}{y-1}\\
&=\dfrac{x(1-x^{m})}{x-1}+\dfrac{y(1-y^{m})}{y-1}\\
&=\dfrac{(y-1)(x-x^{m+1})+(x-1)(y-y^{m+1})}{(x-1)(y-1)}\\
&=\dfrac{(y-1)x-(y-1)x^{m+1}+(x-1)y-(x-1)y^{m+1}}{(x-1)(y-1)}\\
&=\dfrac{2xy-x-y-(y-1)x^{m+1}-(x-1)y^{m+1}}{(x-1)(y-1)}\\
\text{so}\\
0
&=\dfrac{2xy-x-y-(y-1)x^{m+1}-(x-1)y^{m+1}-2(x-1)(y-1)}{(x-1)(y-1)}\\
&=\dfrac{2xy-x-y-2(xy-x-y+1)-(y-1)x^{m+1}-(x-1)y^{m+1}}{(x-1)(y-1)}\\
&=\dfrac{x+y-2-(y-1)x^{m+1}-(x-1)y^{m+1}}{(x-1)(y-1)}\\
\text{or}\\
0
&=x+y-2-(y-1)x^{m+1}-(x-1)y^{m+1}\\
&=(x-1)+(y-1)-(y-1)x^{m+1}-(x-1)y^{m+1}\\
\end{array}
$
Don't know how much this helps,
but here it is.
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