# 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.

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That can't be right. If $r \gt 1$, all of these fractions are positive, but $1/(1-r^2)$ is negative. – joriki Oct 30 '11 at 15:59
In the first line I specify 0 < r < 1. – daniel Oct 31 '11 at 0:51
If r = 0.5 and n = 3, the trig. sum and the ratio (using 3 4 5 4 3 / (3*(1 2 2 2 2 2 1)) both give 0.783069. For n = 2000, the sum is about 1.33316 and the actual value of 1/(1-r^2) is of course 1.333... – daniel Oct 31 '11 at 1:25
I'm sorry, I don't know why I overlooked that. – joriki Oct 31 '11 at 4:32

## 1 Answer

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.

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This is great, thanks. – daniel Oct 31 '11 at 11:19