On an algebraic manipulation of a double summation used to obtain the kth ordinate of the periodogram. I am following Introduction to statistical time series by Fuller.
I am having some problems with what I think is an algebraic manipulation of the double summations in the line where the mouse pointer is at.

If I fix a $k$ how is the mouse pointer line equal to the last result given?
Specifically, what is the meaning of $$\sum_{p= -\infty}^{\infty} \frac{n - |p|}{n}$$ and how does it pop up?
Is a summation $\sum_{p= -\infty}^{\infty}$ still a limit of the partial sums?
Edit: I realized a bit of crucial information is missing:

 A: The expression $\frac{n-|p|}{n}$ doesn't so much "pop up", as appear by design. Presumably the author is wanting to cast the result in terms of the quantity
$$\frac{1}{n-p}\sum_{j=1}^{n-p}(X_j-\mu)(X_{j+p} - \mu)$$
because it is a potential estimator of the autocorrelation function, and although the result is written as an infinite sum, the summands are zero except on a finite set (a square array of points in the $(j,t)$-plane).
The desired result will be established by showing that, for $\omega\ne 0$,
$$\sum_{p =-\infty}^\infty\frac{n-|p|}{n}\tilde{\gamma}(p) \cos \omega p =
 \frac{1}{n}\sum_{t=1}^n\sum_{j=1}^n(X_j-\mu)(X_t - \mu) \cos \omega (t-j)\tag{0*}
$$
where the function $\tilde{\gamma}$ is defined by
$$
\tilde{\gamma}(-p) = \tilde{\gamma}(p) = 
\begin{cases} 
\frac{1}{n-p}\sum_{j=1}^{n-p}(X_j-\mu)(X_{j+p} - \mu) &, 0 \le p\le n-1 \\
0 &, p\gt n-1.
\end{cases}  
$$
Thus, in $(0^*)$, deriving the RHS from the LHS,
$$\begin{align}
& \sum_{p =-\infty}^\infty\frac{n-|p|}{n}\tilde{\gamma}(p) \cos \omega p\tag{1*}\\
& = \gamma(0) + 2 \sum_{p =1}^\infty\frac{n-p}{n}\tilde{\gamma}(p) \cos \omega p\tag{2*}\\
& = \frac{1}{n}\sum_{t=1}^n (X_t-\mu)(X_t - \mu) + \frac{2}{n}\sum_{p =1}^{n-1}\sum_{j=1}^{n-p}(X_j-\mu)(X_{j+p} - \mu) \cos \omega p\tag{3*}\\
&= \frac{1}{n}\sum_{t=1}^n (X_t-\mu)(X_t - \mu) + \frac{2}{n}\sum_{t=1}^n\sum_{j=1}^{t-1}(X_j-\mu)(X_t - \mu) \cos \omega (t-j)\tag{4*}\\
&= \frac{1}{n}\sum_{t=1}^n\sum_{j=1}^n(X_j-\mu)(X_t - \mu) \cos \omega (t-j)\tag{5*}\\
\end{align}
$$
$(2^*)$ is by writing the $p=0$ case separately, and using the given fact that $\tilde{\gamma}(-p) = \tilde{\gamma}(p)$.
$(3^*)$ is by inserting the definitions of $\tilde{\gamma}(0)$ and of $\tilde{\gamma}(p)$ for $p>0$.
$(4^*)$ is by changing the summation variables from $(j,p)$ to $(j,t=j+p)$.
$(5^*)$ is by noting that the total of $(4^*)$ is just the summation over the square array of points $\{(j,t): j\in[1,n], t\in[1,n]\}$, where the first sum is over the main diagonal, and the double sum is over the upper off-diagonals only. Because the whole array of summands is symmetrical about the main diagonal, the total for all the off-diagonals is therefore just twice the latter sum, as in $(4*)$.
(Of course, the above steps can be done in reverse order if we want to show that the LHS can be derived from the RHS.)   
