Using the definition: lim n → ∞ ∑ x = 1 n 1 x − log ⁡ n = γ

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Using the definition:      lim          n      →      ∞            ∑          x      =      1              n            1    x    −  log  ⁡  n  =  γ

hildiadau0o · Accepted Answer

The paper &quot;On the computation of the Euler constant   γ&quot; by Ekatharine A. Karatsuba, in Numerical Algorithms 24(2000) 83-97, has a lot to say about this. This link might work for you.In particular, the author shows that for   k  ≥  1  γ  =  1  −  log  ⁡  k      ∑          r      =      1              12      k      +      1                  (      −      1              )                  r          −          1                            k                  r          +          1                            (      r      −      1      )      !      (      r      +      1      )        +      ∑          r      =      1              12      k      +      1                  (      −      1              )                  r          −          1                            k                  r          +          1                            (      r      −      1      )      !      (      r      +      1              )        2              +      O    (      2          −      k        )and more explicitly                    −                  2                      (            12            k            )            !                          −        2                  k          2                          e                      −            k                          ≤        γ        −        1        +                    log        ⁡        k                  ∑                      r            =            1                                12            k            +            1                                                (            −            1                          )                              r                −                1                                                    k                              r                +                1                                                          (            r            −            1            )            !            (            r            +            1            )                          −                  ∑                      r            =            1                                12            k            +            1                                                (            −            1                          )                              r                −                1                                                    k                              r                +                1                                                          (            r            −            1            )            !            (            r            +            1                          )              2                                                                        ≤                  2                      (            12            k            )            !                          +        2                  k          2                          e                      −            k                              for   k  ≥  1Since the series has fast convergence, you can use these to get good approximations to   γ fairly quickly.

Using the definition: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="M

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