Prove that$$\lim\_{n\to\infty}\frac1n\sum\_{i\_1,i\_2,...i\_k=1}^n\lambda\_1^{|i\_1-i\_2-s\_1|}\lambda\_2^{|i\_2-i\_3-s\_2|}...\lambda\_k^{|i\_k-i\_1-s\_k|}$$is equal to$$\sum\_{j=1}^k\lambda\_j^{S+k-1}\prod\_{l=1,l\ne j}^k\frac{1-\lambda\_l^2}{(\lambda\_j-\lambda\_l)(1-\lambda\_j\lambda\_l)}$$where all $\lambda$s are, in absolute value, smaller than 1, distinct from one another, and $S=|s\_1+s\_2+...+s\_k|$. This can be done by 'brute force' for $k=2$, $k=3$ and $k=4$ but I have no idea how to continue; yet I am reasonably confident that the formula is correct for any $k$.



Prove an existing formula for a limit of a specific sum
Denote $$A=\sum\_{p\_1+\ldots+p\_k=S} \lambda\_1^{|p\_1|}\lambda\_2^{|p\_2|}\ldots
\lambda\_k^{|p\_k|}.\quad (1)$$
We prove two things:


1) Your limit equals $A$. 


2) $A$ equals to what you write.


Start with 1). Note that any summand in your expression is of the form $\lambda\_1^{|p\_1|}\lambda\_2^{|p\_2|}\ldots
\lambda\_k^{|p\_k|}$ for $\sum p\_i=S$ (if $S$ is positive, define $p\_m=i\_{m+1}+s\_m-i\_m$, otherwise $p\_m=i\_m-i\_{m+1}-s\_m$ for $m=1,\ldots,k$, indices are cyclic), and each term appears at most $n$ times, since for fixed $i\_1$ it appears at most once. Thus 
$$
\frac1n\sum\_{i\_1,i\_2,...i\_k=1}^n\lambda\_1^{|i\_1-i\_2-s\_1|}\lambda\_2^{|i\_2-i\_3-s\_2|}...\lambda\_k^{|i\_k-i\_1-s\_k|}\leqslant A,
$$
and the upper limit does not exceed $A$. On the other hand, for any $B<A$ we may find finitely many terms in the sum (1) whose sum exceeds $B$. Each of this terms appears in your sum $n-O(1)$ times, thus we get that the lower limit is at least $B$. Since $B<A$ was arbitrary, it implies that the limit exists and equals to $A$.


2) $A$ is the coefficient of $t^S$ in the Laurent series
$F:=\prod\_i \sum\_{p\in \mathbb{Z}} t^p \lambda\_i^{|p|}$ (which is defined for $|t|$ close enough to 1), $i$-th multiple equals $1/(1-t\lambda\_i)+\lambda\_i/(t-\lambda\_i)=t(1-\lambda\_i^2)/((1-t\lambda\_i)(t-\lambda\_i))$. The product of these guys is a rational function in $t$ which is a linear combinations of the functions $1/(1-\lambda\_jt)$ and $1/(\lambda\_j-t)$:
$$
F=\sum\_j C\_j/(1-\lambda\_jt)+D\_j/(\lambda\_j-t).\quad (2)
$$


$D\_j/(\lambda\_j-t)$ does not provide a coefficient of $t^S$ (it is expressed via negative powers of $t$), the coefficient of $t^S$ may come from $1/(1-\lambda\_jt)=\sum\_{n=0}^\infty \lambda\_j^n t^n$. It remains to calculate the coefficient $C\_j$ in the decomposition (2). We have $C\_j=(F(t)(1-\lambda\_jt))\_{t=1/\lambda\_j}$ that gives your formula.


