Trouble evaluating the sum involving logarithmI was trying to solve this problem: Closed form for ∫ 0 1 log ⁡ log ⁡ ( 1 x + 1 x 2 − 1 ) d xIn the procedure I followed, I came across the following sum: ∑ k = 1 ∞ ( − 1 ) k − 1 k ( ln ⁡ ( 2 k + 1 ) 2 k + 1 − ln ⁡ ( 2 k − 1 ) 2 k − 1 ) I cannot think of any approaches which would help me in evaluating the sum.Any help is appreciated. Thanks!

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Trouble evaluating the sum involving logarithmI was trying to solve this problem: Closed form for       ∫    0    1    log  ⁡  log  ⁡      (          1      x        +                  1                  x          2                    −      1        )        d    xIn the procedure I followed, I came across the following sum:      ∑          k      =      1              ∞        (  −  1      )          k      −      1        k      (                  ln        ⁡        (        2        k        +        1        )                    2        k        +        1              −                  ln        ⁡        (        2        k        −        1        )                    2        k        −        1              )  I cannot think of any approaches which would help me in evaluating the sum.Any help is appreciated. Thanks!

Charlee Beck · Accepted Answer

The terms of the series vanish so we will look only at the partial sums where   n is odd.                                            ∑                      k            =            1                    ∞                (        −        1                  )                      k            −            1                          k                  (                                    log              ⁡              (              2              k              +              1              )                                      2              k              +              1                                −                                    log              ⁡              (              2              k              −              1              )                                      2              k              −              1                                )                                    (1)                                  =                  lim                      n            →            ∞                                    (                      ∑                          k              =              1                        n                    (          −          1                      )                          k              −              1                                k                                    log              ⁡              (              2              k              +              1              )                                      2              k              +              1                                −                      ∑                          k              =              0                                      n              −              1                                (          −          1                      )            k                    (          k          +          1          )                                    log              ⁡              (              2              k              +              1              )                                      2              k              +              1                                )                                    (2)                                  =                  lim                      n            →            ∞                                    (          (          −          1                      )                          n              −              1                                n                                    log              ⁡              (              2              n              +              1              )                                      2              n              +              1                                +                      ∑                          k              =              1                                      n              −              1                                (          −          1                      )                          k              −              1                                log          ⁡          (          2          k          +          1          )          )                                    (3)                                  =                  lim                      n            →            ∞                                    (          (          2          n          +          1          )                                    log              ⁡              (              4              n              +              3              )                                      4              n              +              3                                +                      ∑                          k              =              1                                      n                                log          ⁡                      (                                          4                k                −                1                                            4                k                +                1                                      )                    )                                    (4)                                  =                  lim                      n            →            ∞                          log        ⁡                  (          (          4          n          +          3                      )                                                            2                  n                  +                  1                                                  4                  n                  +                  3                                                              ⋅                      ∏                          k              =              1                                      n                                                          k              −                              1                4                                                    k              +                              1                4                                              )                                    (5)                                  =                  lim                      n            →            ∞                          log        ⁡                  (          2                      n                                              Γ              (              n              +                              3                4                            )                                      Γ              (                              3                4                            )                                                          Γ              (                              5                4                            )                                      Γ              (              n              +                              5                4                            )                                )                                    (6)                                  =        log        ⁡                  (          2                                    Γ              (                              5                4                            )                                      Γ              (                              3                4                            )                                )                                    (7)                                  =        log        ⁡                  (                      1            2                                              Γ              (                              1                4                            )                                      Γ              (                              3                4                            )                                )                    Explanation:  (  1  ): write series as the limit of the partial sums, split into two sums, reindex the second sum  (  2  ): recombine the two sums  (  3  ): substitute   n  ↦  2  n  +  1 and combine pairs of terms of the sum  (  4  ): write a sum of logs as a log of a product  (  5  ):       lim          n      →      ∞            1          2              n              (  4  n  +  3      )                            2          n          +          1                          4          n          +          3                      =  1, write the product as a ratio of Gamma functions  (  6  ): Gautschi&#039;s inequality  (  7  ):   x  Γ  (  x  )  =  Γ  (  x  +  1  )

Cecilia Tapia · Accepted Answer

As said in comments, the sum oscillates between two values and converges to the value   0.391594If we expand in the same spirit as Larsen, what we can see is that  F  (  n  )  =      ∑          k      =      1              n        (  −  1      )          k      −      1        k      (                  ln        ⁡        (        2        k        +        1        )                    2        k        +        1              −                  ln        ⁡        (        2        k        −        1        )                    2        k        −        1              )  can be written as  F  (  n  )  =      ∑          k      =      1              n      −      1        (  −  1      )          k      −      1        log  ⁡  (  2  k  +  1  )  −            (      −      1              )        n            n      log      ⁡      (      2      n      +      1      )              2      n      +      1        Δ  =  (  m  +  1  )      (          1              2        m        +        3              ln    ⁡    (    2    m    +    3    )    −          1              2        m        +        1              ln    ⁡    (    2    m    +    1    )    )  and let us expand it as a Taylor series for large values of   m. We so obtain  Δ  =            log      ⁡              (                  1          m                )            +      1      −      log      ⁡      (      2      )              2      m        +            −      log      ⁡              (                  1          m                )            −      2      +      log      ⁡      (      2      )              2              m        2              +  O      (                  (                  1          m                )            3        )

Trouble evaluating the sum involving logarithm I was trying to solve this problem: Closed form for int_0^1loglog(1/x+sqrt(1/x^2−1)dx

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