Find the limiting value of b_n = a_n+1/ a_n for the Tribonacci sequence a_n+1=a_n+a_n-1+a_n-2 (n >= 2), where a₀=a₁=a₂=1

just like this fibonacci series

6.15. Let On = anti, n > 0, and let L = limn-+09 bn. By definition of the sequence in 6.13., an an+1 an + an-1 an -1 bra = = 1+ = 1+ an an an bn-1 This implies that bobn-1 = bn-1 + 1. Taking the limit as n - co on both sides of the equation, and using the properties of limits, we obtain L' = lim b, lim bm-1 = limb,bn-1 = lim(bn-1 + 1) = limbn-1 + lim1 = [ +1. n n n Hence, the limit satisfies the quadratic equation 12 = L + 1, which is the same equation from Example 6.3. The solution is therefore the Golden Ratio _ = 1+16. 2 Note that this proof was entirely general: Starting with arbitrary positive numbers an and an, if you form the infinite sequence whose next term is the sum of the previous two, i.e., an = an-1 + an-2, n > 2, then the limit of the ratios equals the Golden Ratio: 1+ v5 lim 7-+00 an 2