{ x n } is a sequence and x n + 3 ≤ 1 5 ( x n + 2 + 3 x n + 1 + x n ) for all n ≥ 1. Prove that { x n } is convergent.

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{          x      n        }   is a sequence and       x          n      +      3        ≤      1    5    (      x          n      +      2        +  3      x          n      +      1        +      x    n    ) for all   n  ≥  1. Prove that       {          x      n        }   is convergent.

kejohananws · Accepted Answer

First define      y    n    =  max  (      x    n    ,      x          n      +      1        ,      x          n      +      2        )and show that   (      y    n    ) is a decreasing sequence.Case 1:   (      y    n    ) is not bounded below: Then       y    n    →  −  ∞, which implies that       x    n    →  −  ∞ since       x    n    ≤      y    n  Case 2:   (      y    n    ) is bounded below, and therefore convergent. Let   L  =      lim          n      →      ∞            y    n  . We show that   (      x    n    ) converges to the same value L.Let   ϵ  &amp;gt;  0. Then      x    n    ≤      y    n    &amp;lt;  L  +  ϵfor all sufficiently large   n  ≥  N. It follows that for   n  ≥  N  +  2      x          n      +      1        ≤      1    5    (      x          n      −      2        +  3      x          n      −      1        +            x              n              )  &amp;lt;      1    5    (  4  (  L  +  ϵ  )  +      x    n    )        x          n      +      2        ≤      1    5    (      x          n      −      1        +  2      x          n        +            x              n              +      x          n      +      1        )  &amp;lt;      1    5    (  4  (  L  +  ϵ  )  +      x    n    )        x          n      +      3        ≤      1    5    (            x              n              +  3      x          n      +      1        +      x          n      +      2        )  &amp;lt;      1    5    (  4  (  L  +  ϵ  )  +      x    n    )  and therefore      y          n      +      1        &amp;lt;      1    5    (  4  (  L  +  ϵ  )  +      x    n    )    .But   (      y    n    ) decreases to L, so that  L  &amp;lt;      1    5    (  4  (  L  +  ϵ  )  +      x    n    )    ⟹    L  −  4  ϵ  &amp;lt;      x    n  for   n  ≥  N  +  2We have shown that for all   ϵ  &amp;gt;  0 there is an N such that for   n  ≥  N  +  2  L  −  4  ϵ  &amp;lt;      x    n    ≤  L  +  ϵand that proves       lim          n      →      ∞            x    n    =  L

{ x n </msub> } </mrow> is a sequence and x <mrow cl

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