A series converging (or not) to ln ⁡ 2I have come across the following series, which I suspect converges to ln ⁡ 2 ∑ k = 1 ∞ 1 4 k ( 2 k ) ( 2 k k ) .I could not derive this series from some of the standard expressions for ln ⁡ 2. The sum of the first 100 000 terms agrees with ln ⁡ 2 only up to two digits.Does the series converge to ln ⁡ 2?

Question

A series converging (or not) to   ln  ⁡  2I have come across the following series, which I suspect converges to   ln  ⁡  2      ∑          k      =      1        ∞        1                  4        k            (      2      k      )                          (                            2        k            k                      )              .I could not derive this series from some of the standard expressions for   ln  ⁡  2. The sum of the first   100  000 terms agrees with   ln  ⁡  2 only up to two digits.Does the series converge to   ln  ⁡  2?

Abbigail Vaughn · Accepted Answer

Hint: (or outline -- a lot of details and justifications must be done where there are   (  ⋆  )&#039;s)for   |  x  |  &amp;lt;      1    4  ,                    f        (        x        )                                                                =                                      d              e              f                                                1          2                          ∑                      k            =            1                                ∞                                                              (                                                          2              k                        k                                              )                                                            x            k                    k                =                  1          2                          ∑                      k            =            1                                ∞                                                              (                                                          2              k                        k                                              )                                                ∫          0          x                          t                      k            −            1                          d        t                                                                              =                                      (              ⋆              )                                                1          2                          ∫          0          x                          (                      ∑                          k              =              1                                      ∞                                                                          (                                                                    2                k                            k                                                      )                                                          t                          k              −              1                                )                d        t        =                  1          2                          ∫          0          x                          1          t                          (                      ∑                          k              =              1                                      ∞                                                                          (                                                                    2                k                            k                                                      )                                                          t                          k                                )                d        t                                                                              =                                      (              ⋆              )                                                1          2                          ∫          0          x                          1          t                          (                      1                          1              −              4              t                                −          1          )                d        t                                          =                  1          2                2        ln        ⁡                  2                                    1              −              4              x                        +            1                              &quot;so&quot; (  ⋆)                              (                      ⋆                    )                            f        (        x        )                  →                      t            →                                          1                4                            −                                                ln        ⁡        2

ganolrifv9 · Accepted Answer

Recall that for       |    x      |    &amp;lt;  1 we have      1          1      −              x                  2                      =      ∑          k      =      0              ∞                                    (                                      2          n                n                              )                                x              2        n                    4              n            shifting over the first terms on the series and dividing by x gives      ∑          n      =      1              ∞                                    (                                      2          n                n                              )                                x              2        n        −        1                    4              n              =            1      −              1        −                  x                      2                                      1      −              x                  2                          1    x  Now given an small   ε  &amp;gt;  0 we can integrate both sides from   0 to   1  −  ε, the swap on the RHS is justified due to the uniform convergence, next we let   ε shrink to zero. Wallis product formula yields      1                  4                  n                    2      n                                    (                                      2          n                n                              )                      ∼      1          2      π            1          n              2        n        +        1            Now given an small   ε  &amp;gt;  0 we can integrate both sides from   0 to   1  −  ε, the swap on the RHS is justified due to the uniform convergence, next we let   ε shrink to zero. Wallis product formula yields      1                  4                  n                    2      n                                    (                                      2          n                n                              )                      ∼      1          2      π            1          n              2        n        +        1            which allows us to move in the limit inside and letting   ε  →  0 Hence we have that      ∑          n      =      1              ∞                                    (                                      2          n                n                              )                          1                  4                  n                    2      n        =      ∫          0              1                          1        −                  1          −                      x                          2                                                  1        −                  x                      2                                              d        x            x        =      ∫          0              π              /            2                  1      −      cos      ⁡      (      u      )              sin      ⁡      (      u      )        d  u  =      ∫          0              π              /            2                  sin      ⁡      (      u              /            2      )              cos      ⁡      (      u              /            2      )        d  uIt is pretty obvious what to do next..

I have come across the following series, which I suspect converges to ln2: sum_(k=1)^oo (1)(4^k(2k))((2k),(k)) I could not derive this series from some of the standard expressions for ln2. The sum of the first 100000 terms agrees with ln2 only up to two digits.

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