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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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    The question is what happens to $n^{2H}\varepsilon^2$ and the answer is, this is $n^{2H-2}$ by definition of $\varepsilon$.2012-12-04