Prove that the recurrence f(n+2)=2/3f(n+1)+1/3f(n)f(n+2)=2/3f(n+1)+1/3f(n) for f(0)=5f(0)=5 and f(1)=1f(1)=1 converges

Prove that the recurrence f(n+1)=2/3f(n)+1/3f(n−1)f(n+1)=2/3f(n)+1/3f(n-1) for f(0)=5f(0)=5 and f(1)=1f(1)=1 converges

HINT: derive an expression for f(n)−f(n+1)f(n)-f(n+1)

I dont know exactly what to do here. I must prove this only using the algebra of limits and the theorem that says that a monotonic and bounded sequence converges.

I tried induction over the subsequences f(n+1)=7/9f(n−1)+2/9f(n−2)f(n+1)=7/9 f(n-1)+2/9 f(n-2) to prove monotonicity but I dont get something useful. And the expressions that I derived for f(n)−f(n+1)f(n)-f(n+1) dont give me any clue. Some hint or solution will be appreciated, thank you.

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Relate f(n+1)−f(n)f(n+1) – f(n) and f(n)−f(n−1)f(n) – f(n-1). And think of a geometric …
– Daniel Fischer♦
2 days ago

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Rewrite it to f(n+1)−f(n)=−13(f(n)−f(n−1))f(n + 1) – f(n) = – \frac{1}{3} \bigl( f(n) – f(n – 1) \bigr). Can you go from here?
– Ritz
2 days ago

  

 

Someone cast to close the question! LOL xDDD
– Masacroso
2 days ago

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