Let f ( x , y ) = ∑ n x x 2 + y n 2 . Show that g ( y ) = lim x → ∞ f ( x , y ) exists for all y &gt; 0. Find g ( y ).My first impression of this problem is to use the monotone convergence theorem (MCT) or the dominated convergence theorem (DCT) to interchange the limit and the sum. However, I do not know what to bound the function by. Thanks in advance!If we convert the sum to an integral under the counting measure μ on N , we can re-express the function as f ( x , y ) = ∫ N x x 2 + y n 2 d μIdeally, we want to apply the DCT by finding an integrable function h such that | f n | = | x x 2 + y n 2 | ≤ h a.e. for all n. But, since h has to be an L 1 function independent of n under the counting measure, DCT might not be the correct method.Also, the DCT that I learned involves interchanging ∫ lim n → ∞ f n = lim n → ∞ ∫ f n . But, this problem statement is asking for lim x → ∞ and not lim n → ∞ .

Question

Let   f  (  x  ,  y  )  =      ∑          n            x                  x        2            +      y              n        2            . Show that   g  (  y  )  =      lim          x      →      ∞        f  (  x  ,  y  ) exists for all   y  &amp;gt;  0. Find   g  (  y  ).My first impression of this problem is to use the monotone convergence theorem (MCT) or the dominated convergence theorem (DCT) to interchange the limit and the sum. However, I do not know what to bound the function by. Thanks in advance!If we convert the sum to an integral under the counting measure   μ on       N  , we can re-express the function as  f  (  x  ,  y  )  =      ∫                  N                  x                  x        2            +      y              n        2              d  μIdeally, we want to apply the DCT by finding an integrable function   h such that       |        f    n        |    =      |          x                        x          2                +        y                  n          2                      |    ≤  h a.e. for all   n. But, since   h has to be an       L    1   function independent of   n under the counting measure, DCT might not be the correct method.Also, the DCT that I learned involves interchanging   ∫      lim          n      →      ∞            f    n    =      lim          n      →      ∞        ∫      f    n  . But, this problem statement is asking for       lim          x      →      ∞       and not       lim          n      →      ∞      .

pompatzz8 · Accepted Answer

We can calculate   g  (  y  ) directly instead of interchanging limit and infinite sum:                    g        (        y        )                            =                  lim                      x            →            ∞                                    ∑                      n            =            1                                x                                                x                                          x                                  2                                            +              y                              n                                  2                                                                                                      =                  lim                      x            →            ∞                                                1            x                                    ∑                      n            =            1                                x                                                              x                              2                                                                    x                                  2                                            +              y                              n                                  2                                                                                                      =                  lim                      x            →            ∞                                                1            x                                    ∑                      n            =            1                                x                                                1                          1              +              y              (                              n                x                                            )                                  2                                                                                                      =                  ∫                      0                                1                                                1                          1              +              y                              x                                  2                                                                    d        x                                          =                              1                          y                                                ∫                      0                                1                                                1                          1              +              (                              y                            x                              )                                  2                                                                    d                  y                x                                          =                              1                          y                                      arctan        ⁡                  y                x                                            |                                            0                                1                          =                                            arctan              ⁡                              y                                                    y                                      .

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