Finding the limit for recurrence relation x n + 1 = x n + 1 4 − 1 2

Question

Finding the limit for recurrence relation       x          n      +      1        =            x      n        +          1      4        −      1    2

Jaida Sanders · Accepted Answer

0  &amp;lt;      x          n      +      1        =            x      n        +          1      4        −      1    2    =            x      n                                x          n                +                  1          4                    +              1        2              &amp;lt;      x    n  shows that the sequence is decreasing and bounded below by zero, therefore convergent. The limit L must satisfy  L  =      L    +          1      4        −      1    2  which implies that   L  =  0Then Stolz-Cesaro shows that      lim          n      →      ∞            1          n              x        n              =      lim          n      →      ∞            (          1              x                  n          +          1                      −          1              x        n              )  provided that the latter limit exists. But      1          x              n        +        1              −      1          x      n        =      1                            x          n                +                  1          4                    −              1        2              −      1          x      n        =      1                            x          n                +                  1          4                    +              1        2            converges to one since       x    n   converges to zero.

Finding the limit for recurrence relation x <mrow class="MJX-TeXAtom-ORD"> n

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