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I'm reading the "Time Series: Theory and Methods (2nd ed.)" by P.J.Brockwell and R.A.Davis. I've stopped at the one moment at pp.218-219 (Chapter 7 "Estimation of the mean and the Autocovariance function"). In the proof of theorem 7.1.1 if

$\gamma(n) \rightarrow 0$ as $n \rightarrow +\infty$

then $$lim_{n \rightarrow +\infty} n^{-1} \sum_{|h| < n} \left(|\gamma(h)| \right) = 2 \lim_{n \rightarrow +\infty} \left( |\gamma(n)| \right) = 0$$.

Could anyone explain me the first equality in this part of the proof, pls? I spend much time, but suppose, I'm not so intelligent for self-understanding...((((

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  • $\begingroup$ by the second equality are you referring to $$2 \lim_{n \rightarrow +\infty} \left( |\gamma(n)| \right) = 0$$ this part? if so it's already in your assumption that $y(n)$ approaches 0 as n approches positive infinity $\endgroup$ Commented May 6, 2019 at 18:05
  • $\begingroup$ Sorry, my fault, I've corrected. First equality, of course $\endgroup$ Commented May 6, 2019 at 18:06

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this answer is on hold

first it used the fact that your function $y$ is symmetric around 0 (proof) can be found here, so i don't need to type everything.

then just expanding the summation $$lim_{n \rightarrow +\infty} n^{-1} \sum_{|h| < n} \left(|\gamma(h)| \right) = lim_{n \rightarrow +\infty} n^{-1} * \frac{(y(-n)+y(n))*2n}{2} $$ because h is from -n to n. Then it's pretty self explanatory given that y is symmetric around 0.

$$orig = lim_{n \rightarrow +\infty}y(-n)+y(n) = lim_{n \rightarrow +\infty} 2 y(n)$$

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  • $\begingroup$ Emm... Could you precise explain to me how do you have (y(-n) + y(n))/2 *2n? $\endgroup$ Commented May 6, 2019 at 18:48
  • $\begingroup$ it's just writing out the summation inside the limit, your h is bounded by -n and n, use the arithmetic progression sum formula $\endgroup$ Commented May 6, 2019 at 18:52
  • $\begingroup$ But $gamma(n)$ isn't an arithmetic progression series, I supposed... $\endgroup$ Commented May 6, 2019 at 18:54
  • $\begingroup$ hmm hahaha good point didn't consider that, on first look that was my intuition because of the symmetric properties.. let's wait for a better answer! $\endgroup$ Commented May 6, 2019 at 19:27
  • $\begingroup$ ))) Yeah))) Hope, it also will be so simple - I've spent much time really tried to prove this... $\endgroup$ Commented May 6, 2019 at 19:29

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