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A stationary sequence $(X_n)_{n\in\mathbb{N}}$ exhibits long-range dependence if the autocovariance function $\rho(n):=\mathrm{cov}(X_k,X_{k+n})$ satisfy $$\lim\limits_{n\to\infty}{\rho(n) \over cn^{-\alpha}}=1$$ for some constant $c$ and $\alpha\in (0,1)$.

The increments of the fractional Brownian motion have the long-range dependence property for $H>{1\over2}$ since $$\rho_H(n)={1 \over2}\left[ (n+1)^{2H}+(n-1)^{2H}-2n^{2H}\right]\sim H(2H-1)n^{2H-2}$$ as $n$ goes to infinity.

I cannot follow why this relation holds. Can anybody help please?

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Using $\varepsilon=1/n\to0$, one sees that $$ \rho_H(n)=\tfrac12n^{2H}\left[(1+\varepsilon)^{2H}+(1-\varepsilon)^{2H}-2\right]. $$ Since $(1\pm\varepsilon)^{2H}=1\pm2H\varepsilon+H(2H-1)\varepsilon^2+o(\varepsilon^2)$, $$ \rho_H(n)=\tfrac12n^{2H}\left[1+2H\varepsilon+H(2H-1)\varepsilon^2+1-2H\varepsilon+H(2H-1)\varepsilon^2-2+o(\varepsilon^2)\right], $$ hence $$ \rho_H(n)\sim n^{2H}H(2H-1)\varepsilon^2=H(2H-1)n^{2H-2}. $$

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Maybe it's gonna be a stupid question. I know I can ignore the term $o(\varepsilon^2)$ because it goes to zero as $n\to\infty$. But how can I explain that the term $2H(2H-1)\varepsilon^2$ doesn't disappear due to $\varepsilon^2\to 0$ as $n\to\infty$? – Ichigo Dec 4 '12 at 15:35
The question is what happens to $n^{2H}\varepsilon^2$ and the answer is, this is $n^{2H-2}$ by definition of $\varepsilon$. – Did Dec 4 '12 at 18:28

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