How can this expression be calculated? 1 + 3 ⋯ 4 ⋯ 2 + 5 ⋯ 6 ⋯ I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278... 1 + 3 4 2 + 5 6 ≈ 0.6176 1 + 3 + 7 8 4 + 9 10 2 + 5 + 11 12 6 + 13 14 ≈ 0.6175Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.Is it possible to find a value for this? Will it have a closed form solution?

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How can this expression be calculated?                     1        +                                            3              ⋯                                      4              ⋯                                                  2        +                                            5              ⋯                                      6              ⋯                                          I can&#039;t see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...                    1        +                              3            4                                      2        +                              5            6                                ≈  0.6176                    1        +                                            3              +                                                7                  8                                                                    4              +                                                9                  10                                                                                2        +                                            5              +                                                11                  12                                                                    6              +                                                13                  14                                                                          ≈  0.6175Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.Is it possible to find a value for this? Will it have a closed form solution?

toliwask · Accepted Answer

Define      f    m    (  n  )  =      {                            n          +                                                                                                                                      f                    m                                    (                  2                  n                  +                  1                  )                                                                                                                                                                      f                    m                                    (                  2                  n                  +                  2                  )                                                                                          if                           n              &amp;lt;              m                        ,                                                n                          otherwise.                        Then      f    0    (  0  )  =  0  ,        f    1    (  0  )  =  0  +      1    2    ,        f    2    (  0  )  =  0  +            1      +              3        4              2    ,        f    3    (  0  )  =  0  +            1      +              3        4                    2      +              5        6              ,    …and you&#039;re looking for the value of       lim          m      →      ∞            f    m    (  0  )One can show via reverse induction over   n  =  m  ,  …  ,  1  ,  0 that       f    m    (  n  )  ∈  [  n  ,  n  +  1  ]. So in the limit, defining   f  (  n  )  =      lim          m      →      ∞            f    m    (  n  ), we have   f  (  n  )  ∈  [  n  ,  n  +  1  ]Using interval arithmetic we can then obtain rigorous bounds on   f  (  n  ). Define the interval-valued function  [  f      ]    m    (  n  )  =      {                            n          +                                                                                                                  [                  f                                      ]                    m                                    (                  2                  n                  +                  1                  )                                                                                                                                                  [                  f                                      ]                    m                                    (                  2                  n                  +                  2                  )                                                                                          if                           n              &amp;lt;              m                        ,                                                [          n          ,          n          +          1          ]                          otherwise,                        where the usual interval arithmetic rules apply,  x  +  [  a  ,  b  ]  =  [  x  +  a  ,  x  +  b  ]  ,              [              a        1            ,              b        1            ]              [              a        2            ,              b        2            ]        =      [                  a        1                    b        2              ,                  a        2                    b        1              ]  (because all our intervals are positive, except   [  f      ]    0    (  0  ) which never appears in a denominator). It should be possible to show that       f          k        (  n  )  ∈  [  f      ]    m    (  n  ) for all   k  ≥  m, and so   f  (  n  )  ∈  [  f      ]    m    (  n  ). Assuming that&#039;s true,  [  f      ]          1023        (  0  )  =  [            0.62784196682396      ⏟          542323  ,            0.62784196682396      ⏟          734620  ]narrows down the desired number   f  (  0  ) to   14 significant digits.

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