Limit of sequence of rational polynomials Let $0 < r < 1$. Consider the sequence $S$ of fractions
$$\frac{1}{1+2r+r^2}, \frac{2+2r+2r^2}{2(1+2r+2r^2+2r^3+r^4)},\frac{3+4r+5r^2+4r^3+3r^4}{3(1+2r+2r^2+2r^3+ 2r^4+ 2r^5 +r^6)}, \ldots,\ \text{etc.}  $$
The numerators are built according to the pyramid (the pattern is in the diagonals): 
       1
     2 2 2
   3 4 5 4 3
 4 6 8 8 8 6 4...

and I hope the pattern of the denominator is clear. $S$ seems to converge to 
$\frac{1}{1-r^2}$ as the ratios are built up, and I think I have an outline of a proof. 
It's a matter of showing that the sum $(\frac 1n)\sum_{k=1}^n \frac{1}{1+r^2-2r\cos(\pi k/n)} $ converges to the Poisson integral as $n$ gets large (I am using the version shown in eq. 7 from the pdf below with $a =1, f =1$). 
http://mechse.illinois.edu/research/dstn/teaching_files2/poissonformula.pdf
I think induction might work to show that for each n the sum corresponds to successive elements of S. That part is giving me trouble.   
 A: We can multiply the numerator by $1-r^2$ and check that the resulting fraction converges to $1$ for $0\lt r\lt 1$ (in fact for $|r|\lt 1$).
I'm guessing that by the pattern in the diagonals you mean that the numbers on the diagonals are arithmetic progressions, where the $m$-th "primary" diagonal has initial term $m$ and increment $2m-1$ and the $m$-th "secondary" diagonal has initial term $2m$ and increment $2m$. Then the pyramid would be continued like this:
            1
         2  2  2
      3  4  5  4  3
   4  6  8  8  8  6  4
5  8 11 12 13 12 11  8  5

Then multiplying by $1-r^2$ yields
         1 0 -1
       2 2 0 -2 -2
     3 4 2 0 -2 -4 -3
   4 6 4 2 0 -2 -4 -6 -4
 5 8 6 4 2 0 -2 -4 -6 -8 -5

We can write this as
         2 0 -2
       4 2 0 -2 -4
     6 4 2 0 -2 -4 -6
   8 6 4 2 0 -2 -4 -6 -8
10 8 6 4 2 0 -2 -4 -6 -8 -10

plus single initial and final terms $-n$ and $nr^{2n}$.
Now in the limit $n\to\infty$ (where $n$ is the index of the fraction in the sequence), we can drop all terms that decay with $r^n$, and since the denominator contains a factor of $n$, we can drop all terms in the numerator that don't contain $n$. That simplifies things a lot. In the numerator, we get a constant $-n$ for the initial term, whereas the final term $nr^{2n}$ drops out. For the remaining arithmetic progression we have
$$2\sum_{k=0}^{2n}(n-k)r^k=2\left(n-r\frac{\mathrm d}{\mathrm dr}\right)\frac{1-r^{2n+1}}{1-r}\;.$$
The entire second term drops out because there's nothing in it that contains $n$ but doesn't contain $r^n$, so we're left with
$$-n+2n\frac1{1-r}=n\frac{1+r}{1-r}$$
for the numerator.
The denominator is easier; it's just $n$ times two truncated geometric series in $r$, one starting at $1$ and the the other starting at $r$. Without the $r^n$ terms that arise from the truncation, this is just
$$n\left(\frac1{1-r}+r\frac1{1-r}\right)=n\frac{1+r}{1-r}\;,$$
which confirms your result.
