Simple proof Euler–Mascheroni γ constantI'm searching for a really simple and beautiful proof that the sequence ( u n ) n ∈ N = ∑ k = 1 n 1 k − log ⁡ ( n ) converges.At first I want to know if my answer is OK.My try: lim n → ∞ ( ∑ k = 1 n 1 k − log ⁡ ( n ) ) = lim n → ∞ ( ∑ k = 1 n 1 k + ∑ k = 1 n − 1 [ log ⁡ ( k ) − log ⁡ ( k + 1 ) ] ) = lim n → ∞ ( 1 n + ∑ k = 1 n − 1 [ log ⁡ ( k k + 1 ) + 1 k ] ) = ∑ k = 1 ∞ [ 1 k − log ⁡ ( k + 1 k ) ] Now we prove that the last sum converges by the comparison test: 1 k − log ⁡ ( k + 1 k ) &lt; 1 k 2 ⇔ k &lt; k 2 log ⁡ ( k + 1 k ) + 1which surely holds for k ⩾ 1As ∑ k = 1 ∞ 1 k 2 converges ⇒ ∑ k = 1 ∞ [ 1 k − log ⁡ ( k + 1 k ) ] converges and we name this limit γ

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Simple proof Euler–Mascheroni   γ constantI&#039;m searching for a really simple and beautiful proof that the sequence   (      u    n        )          n      ∈              N              =      ∑          k      =      1        n        1    k    −  log  ⁡  (  n  ) converges.At first I want to know if my answer is OK.My try:      lim          n      →      ∞            (          ∑              k        =        1            n              1      k        −    log    ⁡    (    n    )    )    =      lim          n      →      ∞            (          ∑              k        =        1            n              1      k        +          ∑              k        =        1                    n        −        1              [    log    ⁡    (    k    )    −    log    ⁡    (    k    +    1    )    ]    )    =      lim          n      →      ∞            (          1      n        +          ∑              k        =        1                    n        −        1                    [      log      ⁡      (              k                  k          +          1                    )      +              1        k            ]        )    =      ∑          k      =      1              ∞            [          1      k        −    log    ⁡    (                  k        +        1            k        )    ]  Now we prove that the last sum converges by the comparison test:      1    k    −  log  ⁡  (            k      +      1        k    )  &amp;lt;      1          k      2        ⇔  k  &amp;lt;      k    2    log  ⁡  (            k      +      1        k    )  +  1which surely holds for   k  ⩾  1As       ∑          k      =      1              ∞            1          k      2       converges   ⇒      ∑          k      =      1              ∞            [          1      k        −    log    ⁡    (                  k        +        1            k        )    ]   converges and we name this limit   γ

Kaelyn Barrett · Accepted Answer

One elegant way to show that the sequence converges is to show that it&#039;s both decreasing and bounded below.It&#039;s decreasing because       u    n    −      u          n      −      1        =      1    n    −  log  ⁡  n  +  log  ⁡  (  n  −  1  )  =      1    n    +  log  ⁡  (  1  −      1    n    )  &amp;lt;  0 for all   n. (The inequality is valid because   log  ⁡  (  1  −  x  ) is a concave function, hence lies beneath the line   −  x that is tangent to its graph at   0; plugging in   x  =      1    n   yields   log  ⁡  (  1  −      1    n    )  ≤  −      1    n  .)It&#039;s bounded below because      ∑          j      =      1        n        1    j    &amp;gt;      ∫    1          n      +      1                  d      t        t    =  log  ⁡  (  n  +  1  )  &amp;gt;  log  ⁡  n  ,and so       u    n    &amp;gt;  0 for all   n. (The inequality is valid because the sum is a left-hand endpoint Riemann sum for the integral, and the function       1    t   is decreasing.)

Davian Maynard · Accepted Answer

Upper BoundNote that                              1          n                −        log        ⁡                  (                                    n              +              1                        n                    )                                    =                  ∫          0                      1                          /                        n                                                t                                      d                        t                                1            +            t                                                            ≤                  ∫          0                      1                          /                        n                          t                          d                t                                          =                  1                      2                          n              2                                          Therefore,                    γ                            =                  ∑                      n            =            1                    ∞                          (                      1            n                    −          log          ⁡                      (                                          n                +                1                            n                        )                    )                                                  ≤                  ∑                      n            =            1                    ∞                          1                      2                          n              2                                                                        ≤                  ∑                      n            =            1                    ∞                          1                      2                          n              2                        −                          1              2                                                                        =                  ∑                      n            =            1                    ∞                          1          2                          (                      1                          n              −                              1                2                                              −                      1                          n              +                              1                2                                              )                                                  =        1            Lower BoundNote that                              1          n                −        log        ⁡                  (                                    n              +              1                        n                    )                                    =                  ∫          0                      1                          /                        n                                                t                                      d                        t                                1            +            t                                                            ≥                  ∫          0                      1                          /                        n                                    t                      1            +                          1              n                                                        d                t                                          =                  1                      2            n            (            n            +            1            )                              Therefore,                    γ                            =                  ∑                      n            =            1                    ∞                          (                      1            n                    −          log          ⁡                      (                                          n                +                1                            n                        )                    )                                                  ≥                  ∑                      n            =            1                    ∞                          1                      2            n            (            n            +            1            )                                                            =                  ∑                      n            =            1                    ∞                          1          2                          (                      1            n                    −                      1                          n              +              1                                )                                                  =                  1          2                    A Better Upper BoundUsing Jensen&#039;s Inequality on the concave       t          1      +      t      , we get                              1          n                −        log        ⁡                  (                                    n              +              1                        n                    )                                    =                  1          n                          (          n                      ∫            0                          1                              /                            n                                                          t                                            d                            t                                      1              +              t                                )                                                  ≤                  1          n                                      n                          ∫              0                              1                                  /                                n                                      t                                      d                        t                                1            +            n                          ∫              0                              1                                  /                                n                                      t                                      d                        t                                                            =                  1                      n            (            2            n            +            1            )                              Therefore, since the sum of the Alternating Harmonic Series is   log  ⁡  (  2  )                    γ                            =                  ∑                      n            =            1                    ∞                          (                      1            n                    −          log          ⁡                      (                                          n                +                1                            n                        )                    )                                                  ≤                  ∑                      n            =            1                    ∞                          1                      n            (            2            n            +            1            )                                                            =                  ∑                      n            =            1                    ∞                2                  (                      1                          2              n                                −                      1                          2              n              +              1                                )                                                  =        2        (        1        −        log        ⁡        (        2        )        )

Simple proof Euler–Mascheroni γ constant I'm searching for a really simple and

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