Find the following limits: lim ε → 0 ∑ n = 0 + ∞ ( − 1 ) n 1 + n ϵ lim ε → 0 ∑ n = 0 + ∞ ( − 1 ) n 1 + n 2 ϵ

Question

Find the following limits:       lim          ε      →      0            ∑          n      =      0              +      ∞                  (      −      1              )        n                    1      +      n      ϵ            lim          ε      →      0            ∑          n      =      0              +      ∞                  (      −      1              )        n                    1      +              n                  2                    ϵ

Keyon Fitzgerald · Accepted Answer

Note that      ∑          n      =      0              ∞        (  −  1      )          n            a    n    =      ∑          n      =      0              ∞        (      a          2      n        −      a          2      n      +      1        )  .If       a    n    =  1      /    (  1  +  n  ε  ), then      a          2      n        −      a          2      n      +      1        =      1          1      +      2      n      ε        −      1          1      +      (      2      n      +      1      )      ε        =      ε          (      1      +      2      n      ε      )      (      1      +      (      2      n      +      1      )      ε      )        .Writing   x  =  n  ε, we can replace the sum by an integral:      ∑          n      =      0              ∞                  (      −      1              )                  n                            1      +      n      ε        =      ∑          x      =      0      ,      ε      ,      2      ε      ,      …            ε          (      1      +      2      x      )      (      1      +      2      x      +      ε      )        ∼      ∫          0              ∞                  d      x              (      1      +      2      x              )        2              =      1    2    .Similarly, if       a    n    =  1      /    (  1  +      n    2    ε  ), then      a          2      n        −      a          2      n      +      1        =      1          1      +      (      2      n              )        2            ε        −      1          1      +      (      2      n      +      1              )        2            ε        ∼            4      n      ε              (      1      +      4              n        2            ε              )        2              .Here we take   x  =  n      ε  , giving us      ∑          n      =      0              ∞                  (      −      1              )                  n                            1      +              n        2            ε        =      ∑          x      =      0      ,              ε            ,      2              ε            ,      …                  4      x              ε                    (      1      +      4              x        2                    )        2              ∼      ∫          0              ∞                  4      x      d      x              (      1      +      4              x        2                    )        2              =      1    2  as well.

Find the following limits: <munder> <mo movablelimits="true" form="prefix">lim <mrow c

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