I'm reading Time Series analysis from the book Introduction to time series and forecasting by Peter J. Brockwell and Richard A. Davis. I'm going through the chapter of Spectral Analysis.
Where spectral density defined as $$ f(\lambda) = \frac{1}{2\pi} \sum_{|h| < \infty} \exp{(-ih\lambda)} \gamma(h) $$ Here $\{\gamma(h)\}_{|h|<\infty}$ is is auto covariance function of time series $\{X_t\}_{|t|<\infty}$.
I'm basically stuck in the proof of $f(\lambda)\geq 0 $. there is a step $$ \frac{1}{2\pi N}\sum_{|r|<N}\sum_{|s|<N} \exp{(i\lambda(r-s))}\gamma(r-s) $$ $$ = \frac{1}{2\pi}\sum_{|h|<N}\bigg(1 - \frac{|h|}{N}\bigg) \exp{(i\lambda h)}\gamma(h) $$
How to derive the second line from the first line?
I think this is just a counting argument, but somehow I don't get the result you get in your question, so I'll leave my argument here so maybe it is useful to you or someone can correct it (or you can find a flaw in your book). The summand in the double summation only depends on $r-s$, and when $|r|,|s| < N$, the difference $r-s$ also satisfies $|r-s| < N$. So set $h \overset{def}{=} r-s$ and count the number of pairs $(r,s)$ satisfying $|r| < N$, $|s| < N$ and $r-s = h$.
One easily found such pair is $r=h$ and $s=0$. Since we want $|r-s| = h$ and $|r|$, $|s| < N$, the pairs look like this for say $h > 0$: $$ (-N+1+h, -N+1)), \cdots, (0, -h), \cdots, (h, 0), \cdots, (N-1, N-1-h) $$ Of course, the case $h < 0$ will have the same count, so the answer depends only on $|h|$. You see that the answers $(0,-h), (1,-h+1), \cdots, (h-1,1),(h,0)$ account for $|h|+1$ answers, and for $r < 0$ you have $N-(|h|+1)$ answers, and similarly for the answers with $s > 0$. This gives you a total of $2(N-(|h|+1)) + |h|+1 = 2N-(|h|+1)$ such pairs. So you can re-write your sum as $$ \frac 1{2\pi N} \sum_{|r|,|s| < N} \mathrm{exp}(i\lambda(r-s)) \gamma(r-s) $$ by putting together the terms involving $(r,s)$ with $|r|,|s| < N$ and $r-s = h$ and just put your count in front of the term, which gives $$ \frac 1{2\pi N} \sum_{|h| < N} (2N-(h+1)) \mathrm{exp}(i\lambda h) \gamma(h) $$ By bringing the factor $\frac 1N$ in the summation, you get $$ \frac 1{2\pi} \sum_{|h| < N} \left( 2-\frac{|h|+1}N \right) \mathrm{exp}(i\lambda h) \gamma(h) $$ One possible reason that I see for the discrepancy in our counts is if you're really summing over $0 \le r,s < N$ instead of $|r|,|s| < N$, because then the count becomes $N-|h|$ instead of $2N-(|h|+1)$, and you get the correct factor. That may explain your mistake (or the author's).
Hope that helps,