Series for logarithmsThis is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues. ∑ n = 0 ∞ ( 1 2 n + 1 − 1 2 n + 2 ) = ln ⁡ 2 ∑ n = 0 ∞ ( 1 3 n + 1 + 1 3 n + 2 − 2 3 n + 3 ) = ln ⁡ 3 ∑ n = 0 ∞ ( 1 4 n + 1 + 1 4 n + 2 + 1 4 n + 3 − 3 4 n + 4 ) = ln ⁡ 4 e t c .

Question

Series for logarithmsThis is more of a challenge than a question, but I thought I&#039;d share anyway. Prove the following identities, and prove that the pattern continues.      ∑          n      =      0        ∞        (          1              2        n        +        1              −          1              2        n        +        2              )    =  ln  ⁡  2      ∑          n      =      0        ∞        (          1              3        n        +        1              +          1              3        n        +        2              −          2              3        n        +        3              )    =  ln  ⁡  3      ∑          n      =      0        ∞        (          1              4        n        +        1              +          1              4        n        +        2              +          1              4        n        +        3              −          3              4        n        +        4              )    =  ln  ⁡  4      e    t    c    .

Dayana Powers · Accepted Answer

Let   m  ≥  2. Put:      S    m    =      ∑          n      ≥      0        (      1          m      n      +      1        +  ⋯  +      1          m      n      +      m      −      1        −            m      −      1              m      n      +      m        )As             1              m        n        +        r              =          ∫      0      1              x              m        n        +        r        −        1              d    x  , We have      S    m    =      ∫    0          1                  1      +      x      +      ⋯      +              x                  m          −          2                    −      (      m      −      1      )              x                  m          −          1                            1      −              x        m              d  xBut  1  +  x  +  ⋯  +      x          m      −      2        −  (  m  −  1  )      x          m      −      1        =  (  1  −      x          m      −      1        )  +  ⋯  +  (      x          m      −      2        −      x          m      −      1        )Hence  1  +  x  +  ⋯  +      x          m      −      2        −  (  m  −  1  )      x          m      −      1        =  [  (  1  −  x  )  (  1  +  x  +  ⋯      x          m      −      2        )  ]  +  ⋯  +  [      x          m      −      2        (  1  −  x  )  ]and  [  (  1  +  x  +  ⋯      x          m      −      2        )  ]  +  ⋯  +  [      x          m      −      2        ]  =  1  +  2  x  +  ⋯  +  (  m  −  1  )      x          m      −      2      Thus      S    m    =      ∫    0    1                      ∑                  k          =          0                          m          −          2                    (      k      +      1      )              x                  k                            1      +      x      +      ⋯      +              x                  m          −          1                      d  x  =      ∫    0    1                      P                  ′                    (      x      )              P      (      x      )        d  x  =  [  log  ⁡  (  1  +  x  +  ⋯  +      x          m      −      1        )      ]    0    1    =  log  ⁡  (  m  )

firmezas1 · Accepted Answer

For any   m  ∈      N    ,      ∑          k      =      0              ∞                  (            1          m      k      +      1        +      1          m      k      +      2        +  .  .  .  +      1          m      k      +      m      −      1        −            m      −      1              m      k      +      m                  )        =      lim          n      →      ∞                  (            ∑          k      =      0              m      n            1          m      k      +      1        +  .  .  .  +      1          m      k      +      m      −      1        −            m      −      1              m      k      +      m                  )        =      lim          n      →      ∞                  (            ∑          k      =      1              m      n            1    k    −  m      ∑          k      =      1        n        1          m      k                  )        =      lim          n      →      ∞                  (            ∑          k      =      1              m      n            1    k    −      ∑          k      =      1        n        1    k              )        =      lim          n      →      ∞                  (            ∑          k      =      1              m      n            1    k    −  ln  ⁡  (  m  n  )  +  ln  ⁡  (  m  n  )  −      ∑          k      =      1        n        1    k              )        =      lim          n      →      ∞                  (            ∑          k      =      1              m      n            1    k    −  ln  ⁡  (  m  n  )  +  ln  ⁡  (  n  )  −      ∑          k      =      1        n        1    k              )        +  ln  ⁡  (  m  )  =  γ  −  γ  +  ln  ⁡  (  m  )  =  ln  ⁡  (  m  )  ,where   γ is the Euler-Mascheroni constant.

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