Given is the sequence x 1 = 0 , x n + 1 = 2 + x n . Prove: lim n → ∞ 2 n 2 − x n = πHint:Use the following formulas: cos ⁡ ( x 2 ) = 1 + cos ⁡ x 2 sin ⁡ ( x 2 ) = 1 − cos ⁡ x 2 Any idea how to solve this problem?

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Given is the sequence       x    1    =  0  ,        x          n      +      1        =      2    +          x      n      . Prove:      lim          n      →      ∞            2    n        2    −          x      n        =  πHint:Use the following formulas:  cos  ⁡      (          x      2        )    =                    1        +        cos        ⁡        x            2        sin  ⁡      (          x      2        )    =                    1        −        cos        ⁡        x            2      Any idea how to solve this problem?

bluayu0y · Accepted Answer

Note that as       x    1    ≤  2,       x    n    ≤  2  ∀  n. This can be shown inductively.Thus, we may write       x    n    =  2  cos  ⁡      θ    n   for some       θ    n    ∈  [  0  ,  π      /    2  ], with       θ    1    =  π      /    2Now,   2  cos  ⁡      θ          n      +      1        =      2    (    1    +    cos    ⁡          θ      n        )    =  2  cos  ⁡            θ      n        2      ⟹        θ          n      +      1        =            θ      n        2    =            θ      1              2      n        =      π          2              n        +        1            This implies that       2    n        2    −          x      n        =      2          n      +      1        sin  ⁡      π          2              n        +        1            . Finish off by using             sin      ⁡      x        x    →  1 as   x  →  0

meindwrhc · Accepted Answer

By induction and the formula from the hint, we can easily get       x    n    =  2  cos  ⁡  (      π          2      n        ). Then      lim          n      →      ∞            2    n        2    −          x      n        =      lim          n      →      ∞            2    n        2        1    −    cos    ⁡    (          π              2        n              )    =      lim          n      →      ∞            2          n      +      1        sin  ⁡      (          π              2                  n          +          1                      )    =  π

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