Let u n and v n be two sequences defined by 0 &lt; v 1 &lt; u 1 , u n + 1 = u n 2 + v n 2 u n + v n , v n + 1 = u n + v n 2 (a) Show that u n and v n are monotonic sequences.(b) Show that u n and v n converge to the same limit.

Question

Let       u    n   and       v    n   be two sequences defined by  0  &amp;lt;      v    1    &amp;lt;      u    1  ,       u          n      +      1        =                    u        n        2            +              v        n        2                            u        n            +              v        n            ,       v          n      +      1        =                    u        n            +              v        n              2  (a) Show that       u    n   and       v    n   are monotonic sequences.(b) Show that       u    n   and       v    n   converge to the same limit.

trantegisis · Accepted Answer

Let’s see the relation between       u          n      +      1       and       v          n      +      1      Consider,      u          n      +      1        −      v          n      +      1        =                    u        n        2            +              v        n        2                            u        n            +              v        n              −                    u        n            +              v        n              2    =            2              u        n        2            +      2              v        n        2            −      (              u        n        2            +              v        n        2            +      2              u        n                    v        n            )              2      (              u        n            +              v        n            )        =                    u        n        2            +              v        n        2            −      2              u        n                    v        n                    2      (              u        n            +              v        n            )        =            (              u        n            −              v        n                    )        2                    2      (              u        n            −              v        n            )      For   n  =  1      u    2    −      v    2    =            (              u        1            −              v        1                    )        2                    2      (              u        1            +              v        1            )        &amp;gt;  0 (because       u    1    &amp;gt;  0  ,      v    1    &amp;gt;  0)      u    2    &amp;gt;      v    2  Suppose that       u          k      +      1        &amp;gt;      v          k      +      1       is true for   k  =  nAs the numerator and denominator are both positive, we have       u          k      +      2        −      v          k      +      2        &amp;gt;  0 for   n  =  k  +  1This principle must be true for all   n  ∈  Na) Look at the sequence (      v    n  )Consider,      v          n      +      1        −      v    n    =                    u        n            +              v        n              2    −      v    n    =                    u        n            −              v        n              2    &amp;gt;  0 (as       u    n    &amp;gt;      v    n  , for all   n  ∈  N)Thus, the sequence is monotonically increasing And this sequence is bounded below by 0. (by the given data)2) Look at the second sequence      u          n      +      1        −      u    n    =                    u        n        2            +              v        n        2                            u        n            +              v        n              −      u    n    =                    u        n        2            +              v        n        2            −              u        n        2            −              u        n                    v        n                            u        n            +              v        n              =                    v        n            (              v        n            −              u        n            )              (              v        n            +              u        n            )            v    n    &amp;gt;  0  ,  ∀  n  ∈  N and       v    n    −      u    n    &amp;lt;  0  ,  ∀  n  ∈  NThus,       v    n    +      u    n    &amp;gt;  0  ,  ∀  n  ∈  N  because   0  &amp;lt;      v    n    &amp;lt;      u    n    ,  ∀  n  ∈  NTherefore, we have       u          n      +      1        −      u    n    &amp;lt;  0  ,  ∀  n  ∈  NHence, this sequence is monotonically decreasingAs       v    n    &amp;lt;      u    n    ,  ∀  n  ∈  N, we can write that       v    n    &amp;lt;      u    1  , ensuring that       v    n   is bounded above,       v    n   is increasing. By Monotone Convergence Theorem, the sequence   (      v    n    ) is convergent.Similarly,      u    n   is bounded below by 0 and       u    n   is decreasing , we must have       u    n   to be convergent.Let       lim          n      →      ∞            v    n    =  m so that       lim          n      →      ∞            v          n      +      1        =  mAnd       lim          n      →      ∞            u          n        =  n so that       lim          n      →      ∞            u          n      +      1        =  nUse       v          n      +      1        =                    u        n            +              v        n              2   in the equation       u          n      +      1        =                    u        n        2            +              v        n        2                            u        n            +              v        n            Thus, we get      u          n      +      1        =                    u        n        2            +              v        n        2                            2        (                  u          n                +                  v          n                )            2            u          n      +      1        =                    u        n        2            +              v        n        2                    2              v                  n          +          1                      2      u          n      +      1            v          n      +      1        =      u    n    2    +      v    n    2        lim          n      →      ∞        (  2      u          n      +      1            v          n      +      1        )  =      lim          n      →      ∞        (      u    n    2    +      v    n    2    )  2  n  m  =      n    2    +      m    2        n    2    +      m    2    −  2  n  m  =  0  (  n  −  m      )    2    =  0  n  =  mThus, these sequences converge to the same limit

Let u n </msub> and v n </msub> be two sequences defined by

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