Let s n = ∑ j = 1 n ∑ i = n + 1 + ∞ 1 i 2 + j 2

Question

Let      s    n    =      ∑          j      =      1        n        ∑          i      =      n      +      1              +      ∞                  1                        i          2                +                  j          2

Scarlet Reid · Accepted Answer

Let  f  (  x  ,  y  )  =      1                  x        2            +              y        2              .The series       s    n   is the Riemann sum approximation to the integral  G  =      ∫    0    1        ∫          1        ∞    f  (  x  ,  y  )  d  y  d  x  .with the sum      s    n    =      ∑          i      =      1        n        ∑          j      =      n      +      1              ∞            1          n      2        f      (          i      n        ,          j      n        )  In each small square   [  (  i  −  1  )      /    n  ,  i      /    n  ]  ×  [  (  j  −  1      /    n  ,  j      /    n  )  ], we have the first order approximation      ∫          (      i      −      1      )              /            n              i              /            n            ∫          (      j      −      1      )              /            n              j              /            n        f  (  x  ,  y  )  d  y  d  x  −      1          n      2        f      (          i      n        ,          j      n        )    =      ∫          (      i      −      1      )              /            n              i              /            n            ∫          (      j      −      1      )              /            n              j              /            n            ∂    x    f      (          i      n        ,          j      n        )        (    x    −          i      n        )    +      ∂    y    f      (          i      n        ,          j      n        )        (    y    −          i      n        )    d  y  d  x  +  O  (      n          −      4        )  =  −      1          2              n        3              (      ∂    x    f      (          i      n        ,          j      n        )    +      ∂    y    f      (          i      n        ,          j      n        )    )  +  O  (      n          −      4        )  .Thus we conclude that      lim          n      →      ∞        n  (  G  −      s    n    )  =  −      1    2        ∫    0    1        ∫          1        ∞        ∂    x    f  (  x  ,  y  )  +      ∂    y    f  (  x  ,  y  )  d  y  d  x  .By FTC, we can easily evaluate the integral to be -1, as      ∫    0    1        ∫          1        ∞        ∂    x    f  (  x  ,  y  )  d  y  d  x  =      ∫          1        ∞    f  (  1  ,  y  )  −  f  (  0  ,  y  )  d  y  =      π    4    −  1  ,      ∫    0    1        ∫          1        ∞        ∂    y    f  (  x  ,  y  )  d  y  d  x  =      ∫          0        1    −  f  (  x  ,  1  )  d  x  =  −      π    4    .So we have the desired result.

Let s n </msub> = <munderover> ∑ <mrow class="MJX-TeXAto

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