Break a finite double sum $\sum_{i,j=1}^{n}$ into $|i-j|K$. This question is about a proof from Amir Dombo's note. 
In the proof: he break $\sum_{i,j=1}^{n}Cov(X_{i}, X_{j})$ into $\{(i,j):|i-j|<K\}$ and $\{(i,j):|i-j|\geq K\}$, so that he can bounde the whole sum, using the pre-proved facts that $$(1)\ Cov(X_{i}, X_{j})\leq\epsilon\ \text{for}\ |i-j|\geq K,$$ and $$(2)\ Cov(X_{i}, X_{j})\leq r(0)\ \text{for all}\ i,j,$$ where $r(0)$ is a constant.
The bound he got is $$Cov(X_{i}, X_{j})\leq 2Knr(0)+n^{2}\epsilon,$$ I understand how he got $n^{2}\epsilon$, as follows: 
$$\sum_{i,j=1}^{n}Cov(X_{i}, X_{j})=\sum_{|i-j|<K}Cov(X_{i}, X_{j})+\sum_{|i-j|\geq K}Cov(X_{i}, X_{j}),$$ and for the second sum, we use the fact $(1)$, $$\sum_{|i-j|\geq K}Cov(X_{i}, X_{j})\leq\sum_{|i-j|\geq K}\epsilon\leq\sum_{i,j=1}^{n}\epsilon=n^{2}\epsilon,\  \text{since}\ \epsilon>0.$$
However, the same technique cannot be applied to $\sum_{|i-j|<K}Cov(X_{i}, X_{j})$, since we $r(0)$ is not necessarily positive.
What should I do to evaluate $\sum_{|i-j|<K}Cov(X_{i}, X_{j})$? or to bound it?
Thank you!
Edit 1:
I came up with something which is close to the bound, but I did not know what is happening... I firstly tried this:
$$\sum_{|i-j|<K}r(0)=\sum_{j-K<i<K+j}r(0)=\sum_{i=j-K+1}^{K+j-1}r(0)=(2K-1)r(0),$$ so it seems that we are losing a factor of $n$, so if I use this $$\sum_{|i-j|<K}r(0)=\sum_{j=1}^{n}\sum_{j-K<i<K+j}r(0)=\sum_{j=1}^{n}\sum_{i=j-K+1}^{K+j-1}r(0)=n(2K-1)r(0),$$ which is close to the bound since the bound is $2nKr(0)$.
Also, what I did is the reverse engineering. If I don't know the answer in the first place, I will not produce such an answer (which is just close)
Also, does the $\sum_{j-K<i<K+j}$ over-count? What if $K>j$ then $i$ will sum starting at a negative number.
Is there a way to produce $\sum_{|i-j|<K}=\sum_{j=1}^{n}\sum_{i=j-K+1}^{K+j}$ rigorously?
Is there a general way to decompose the double sum?
 A: Specially when bounding the variance of partial sums from dependent sequence of random variables, it's usual to distinguish covariances of short and long lag lengths.
It's clear that $A:=\{(i,j):\lvert i-j\rvert<K ,\quad 1\leq i,j\leq n\}=\cup_{l=1}^n\{(l,j):\lvert l-j\rvert<K, \quad 1\leq j\leq n\}:=A_l$. In addition, the colletion $\{A_l\}_{l=1}^n$ is clearly pairwise disjoint. It means that the cardinality of $A$ is the sum of the cardinality of the $A_l's$. Now focus on $A_l$.
Assume $K\in \mathbb{N}$. Denote $[n]=\{1,2,\dotsc,n\}$ and $B_K(l)=\{j\in\mathbb{R}:\lvert l-j\rvert<K\}$, a open ball centered at $l$ with radius $K$, on the usual metric space. Note that $A_l=(\mathbb{R}\cap[n])\cap B_K(l)=[n]\cap B_K(l)$. That ($\mathbb{R}\cap[n]$) is just  the line/section in the plane when the first coordinate is fixed to $l$, intersecting the naturals $[n]$. Which naturals in $[n]$ are in the open ball $B_K(l)$? Instead of seek the exact result for all cases satisfying $n>K$, note that for each $l$, there are four possibilities


*

*$\nexists K-1$ indices smaller than $l$ and $\exists K-1$ indices greater that $l$;

*$\exists K-1$ indices smaller than $l$ and $\nexists K-1$ indices greater that $l$;

*$\nexists K-1$ indices smaller than $l$ and $\nexists K-1$ indices greater that $l$;

*$\exists K-1$ indices smaller than $l$ and $\exists K-1$ indices greater that $l$;;


Obviously, $\#A_l=2 K -1$ with $l$ under case $4$ is the maximum along $l\in [n]$.Hence $$\#A\leq n(2 K -1)\leq n2 K$$
So, basically this is the result that you got, with the exception of the last inequality.
This implies the bound for the sum of covariances that he got. 
But if $K\in\mathbb{R}$, more work is needed since we cannot use  $\#A\leq n(2\left \lceil{K}\right \rceil -1)$ to obtain the bound $n2 K$. If you are primarily worried with the asymptotics, then it doesn't matter. The covariances' sum for the short lag case is $O(n)$, and the (dominant term) long lag case is $O(n^2)$.
Comment
Now that I read your comment about dividing the sum by $n^2$, you can simpy use the bound $n(2\left \lceil{K}\right \rceil-1)$ since it's $o(n^2)$. To obtain the convergence you need to allow $\epsilon$ to be arbitrarily small though.
A: Today,  I came across a similar problem as yours. I did that differently though in 2 and a half lines (see Claim 7):

If I allow the distances to be zero, the bound for $A$ still holds.
