Let $n$ be postive integer number, and $x_{i}\ge 0$, such 
$$x_{i}x_{j}\le 4^{-|i-j|},1\le i,j\le n$$
then I have prove 
$$x_{1}+x_{2}+\cdots+x_{n}<\dfrac{5}{3}$$
**Edit Add Proof**:since $x^2_{i}\le 1,0\le x_{i}\le 1$,Let $S_{j}=\sum_{i=1}^{j}x_{i},S=\sum_{i=1}^{n}x_{i}$,then we have
$$0=S_{0}\le S_{1}\le S_{2}\le\cdots\le S$$so there exist $k$,such
$S_{k}\le\dfrac{S}{2}\le S_{k+1}$,
if we let $$T_{k}=S-S_{k},T_{k+1}=S-S_{k+1}$$
then we have
$$|S_{k}-T_{k}|+|S_{k+1}-T_{k+1}|=|2S_{k}-S|+|2S_{k+1}-S|=2x_{k+1}\le 2$$
then for $l\in\{k,k+1\}$,we have
$$|S_{l}-T_{l}|\le 1\tag{1}$$
and we have
$$S_{l}T_{l}=\sum_{i=1}^{l}\sum_{j=l+1}^{n}x_{i}x_{j}\le\sum_{i=1}^{l}\sum_{j=l+1}^{n}4^{-|i-j|}\le\sum_{i=1}^{l}\dfrac{1}{4^{l-i}}\sum_{j=l+1}^{n}\dfrac{1}{4^{j-l}}<\dfrac{4}{3}\cdot\dfrac{1}{3}\tag{2}$$
use $(1)$ and $(2)$
we have
$$x_{1}+x_{2}+\cdots+x_{n}=S_{l}+T_{l}=\sqrt{(S_{l}-T_{l})^2+4S_{l}T_{l}}\le\dfrac{5}{3}$$


**Question :**

Let $n$ be postive integer number, and $x_{i}\ge 0$, such 
$$x_{i}x_{j}x_{k}\le 4^{-|i-j-k|},1\le i,j,k\le n$$
then I have prove 
$$x_{1}+x_{2}+\cdots+x_{n}<C$$
find the best constant  $C?$