Limit of $\lim_{p \rightarrow \infty} \frac{1}{p}\sum_{i=1}^{p} \sum_{j=i+1}^{p} \frac{c^2 r^{j-i}}{\sqrt{c^2 - S_i^2}\sqrt{c^2 - S_j^2} }$ 
Assume that $r \in (0,1)$, I'm looking for the limit $$A:= \lim_{p \rightarrow \infty} \frac{1}{p}\sum_{i=1}^{p} \sum_{j=i+1}^{p}  \frac{c^2 
 r^{j-i}}{\sqrt{c^2 - S_i^2}\sqrt{c^2 - S_j^2} } $$
where $S_i = \sum_{u=1}^{p}\frac{r^{|u-i|}}{u}$ and $c^2 > S_j^2$ is a known constant.

I'm able to show that $$ \sum_{i=1}^{p} \sum_{j=i+1}^{p}\frac{c^2 r^{j-i}}{p} \to_{p \to \infty} \frac{c^2 r }{1 - r}$$
Therefore, one idea is to find $j^*$ (or just $j^*=1$, which should still be quite rough), for which $S_{j^*} > S_j$ and get an upper bound by plugging it in instead of $S_i, S_j$, thus managing the calculations a bit. This suggests that the sum should be converging, however, I'm not yet confident in how to properly handle the squared-roots in the denominator.
On the other hand, $S_j \rightarrow 0$, therefore for $i,j \gg N$, we're summing $\approx \frac{c^2 r^{j-i}}{\sqrt{c^2 - \varepsilon} \sqrt{c^2 - \varepsilon}  } \approx 1 \cdot r^{j-i}$. However, all the $i,j < N$ should amount to something significant.
Lastly, maybe a Taylor expansion of the squared-root term could help, however, I'm not really sure it's that straightforward when applied on series.
My question:
What could be the possible alternative ways to rewrite the squared-root terms in the denominator? Are there any other known good tricks that would work here?
 A: To make your argument [leading to $A=r/(1-r)$] rigorous, let's use the Stolz–Cesàro theorem (SCT; more specifically, this application) and the dominated convergence theorem (DCT; actually, the discrete version of it, also known as Tannery's theorem).
Some rearrangement first. Let $c_{i,j}=c^2(c^2-S_i^2)^{-1/2}(c^2-S_j^2)^{-1/2}$ for brevity. Then $$\sum_{i=1}^{p\color{red}{-1}}\sum_{j=i+1}^{p}c_{i,j}r^{j-i}=\sum_{i=1}^{p-1}\sum_{k=1}^{p-i}c_{i,i+k}r^k=\sum_{k=1}^{p-1}r^k\sum_{i=1}^{p-k}c_{i,i+k}\\\implies\color{blue}{A=\lim_{p\to\infty}\sum_{k=1}^{p-1}b_{p,k}r^k}\quad\text{where}\quad b_{p,k}=\frac1p\sum_{i=1}^{p-k}c_{i,i+k}.$$
As you noted, $\lim\limits_{i\to\infty}S_i=0$, thus $\lim\limits_{i\to\infty}c_{i,i+k}=1$ for each (fixed) $k$, and then $\lim\limits_{p\to\infty}b_{p,k}=1$ by SCT.
It remains to (be able to) apply DCT to the $\color{blue}{\text{blue}}$ equality. But $S_j\to 0$ implies that $s:=\max_j S_j$ exists, and we're given $c>S_j$ for all $j$, hence $c>s$. So $c_{i,j}\leqslant M:=c^2/(c^2-s^2)$ and $b_{p,k}\leqslant M$. This gives a dominating (geometric) series for DCT, and we get $A=\sum_{k=1}^\infty r^k=r/(1-r)$.
