Am asked to show that $$\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n \rho^{|i-j|}\to \frac{1+\rho}{1-\rho} \quad \text{as} \quad n\to\infty$$ where $\rho$ is a number satisfying $|\rho|<1$.
My attempt:
Think of an $n\times n$ matrix having coefficient $\rho^{|i-j|}$ in position $(i,j)$. The coefficients are constant along the diagonals $|i-j|=k$ for $k=0,\dots, n-1$. Summing across these diagonals we get $$\sum_{i=1}^n\sum_{j=1}^n \rho^{|i-j|}=2\sum_{i=0}^{n-1} (n-i)\rho^i -n$$ and so $$\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n \rho^{|i-j|}=2\sum_{i=0}^{n-1} \rho^i-\frac{2}{n}\sum_{i=0}^{n-1} i\rho^i -1$$ The first term on the RHS converges to $\frac{2}{1-\rho}$. For the second term we use l'Hospital's rule to get
$$ \lim_{x\to \infty} x\rho^x=\lim_{x\to \infty} \frac{x}{e^{-x\log(\rho)}}=\lim_{x\to \infty} \frac{-1}{\log(\rho)e^{-x\log(\rho^)}}=0$$
and so the using the result from here we get that the second term converges to zero. The conclusion follows.
Am I missing something?
Here is another approach. Note that we can write
$$\begin{align} \frac1n\sum_{i=1}^n\sum_{j=1}^n \rho^{|i-j|}&=1+\frac2n\sum_{i=1}^n\sum_{j=1}^{i-1}\rho^{i-j}\\\\ &=1+\frac2n\sum_{i=1}^n \rho^i \sum_{j=1}^{i-1}\left(\frac1{\rho}\right)^j\\\\ &=1+\frac2n\sum_{i=1}^n \frac{\rho-\rho^i}{1-\rho}\\\\ &=\frac{1+\rho}{1-\rho}-\frac2{n(1-\rho)} \sum_{i=1}^n \rho^{i} \end{align}$$
Taking the limit as $n\to \infty$ we find that
$$\begin{align} \frac1n\sum_{i=1}^n\sum_{j=1}^n \rho^{|i-j|}&=\frac{1+\rho}{1-\rho} \end{align}$$
as was to be shown!