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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?

Alphie
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  • You are right. The sum $\sum_{i=0}^{n-1}i \rho^i$ can also be computed using the derivative of $\rho \mapsto \sum_{i=0}^{n-1} \rho^i$. – Christophe Leuridan Nov 30 '22 at 21:09
  • @ChristopheLeuridan How? If $f(\rho)=\sum_{i=0}^{n-1} \rho^i$ then $\sum_{i=0}^{n-1}i \rho^i=\rho f'(\rho)$. Can it be more explicit? – Alphie Nov 30 '22 at 22:55
  • Of course, because $f(\rho) = (1-\rho^n)/(1-\rho)$ whenever $\rho \ne 1$. – Christophe Leuridan Dec 01 '22 at 07:48
  • @ChristopheLeuridan Ah yes I get $\sum_{i=0}^{n-1}i \rho^i=\frac{1-\rho^n}{(1-\rho)^2}-\frac{n\rho^{n-1}}{1-\rho} \to \frac{1}{(1-\rho)^2}$ – Alphie Dec 01 '22 at 15:36

1 Answers1

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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!

Mark Viola
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