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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 $k=2$ and $k=3$, but I have no idea how to continue; yet I am reasonably confident that the formula is correct for any $k$.

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 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 $k=2$ and $k=3$, but I have no idea how to continue; yet I am reasonably confident that the formula is correct for any $k$.

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 $k=2$ and $k=3$, but I have no idea how to continue; yet I am reasonably confident that the formula is correct for any $k$.

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 $k=1$ and $k=2$, but I have no idea how to continue; yet I am reasonably confident that the formula is correct for any $k$.

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 $k=2$ and $k=3$, but I have no idea how to continue; yet I am reasonably confident that the formula is correct for any $k$.