How to show 1 2 ⋅ 3 4 ⋅ 5 6 ⋯ 99 100 &lt; 1 12 ?

Question

How to show       1    2    ⋅      3    4    ⋅      5    6    ⋯      99    100    &amp;lt;      1    12  ?

relientaaho2 · Accepted Answer

For   m  ,  n  ∈      N   with   m  ≤  n, let       S          m      ,      n        :=      ∏          i      =      m        n                2      i      −      1              2      i      , then       S          m      ,      n        2    =      ∏          i      =      m        n                (                        2          i          −          1                          2          i                    )        2    ≤      ∏          i      =      m        n          (                  2        i        −        1                    2        i              )        (                  2        i                    2        i        +        1              )    =      ∏          i      =      2      m      −      1              2      n              i          i      +      1        =            2      m      −      1              2      n      +      1          .Therefore,       S          m      ,      n        ≤                    2        m        −        1                    2        n        +        1            . In particular,       S          4      ,      50        ≤            7      101      . Now,       S          1      ,      50        =      1    2    ⋅      3    4    ⋅      5    6    ⋅      S          4      ,      50      . Hence      S          1      ,      50        ≤      15    48              7      101        &amp;lt;      1    12      .You can also show that       S          m      ,      n        ≥                    m        −        1            n      , so       S          1      ,      50        ≥      1    2    ⋅      3    4    ⋅      5    6    ⋅      7    8    ⋅            4      50        =      7          64              2            . This will give you       1    13    &amp;lt;      S          1      ,      50        &amp;lt;      1    12  . Asymptotically,       S          1      ,      n        ∼      1          π      n      . Indeed,      1          π              (        n        +                  1          2                )              &amp;lt;      S          1      ,      n        &amp;lt;      1          π      n          .

istupilo8k · Accepted Answer

Assuming that you know factorials and their approximations, rewrite  A  =      1    2    ×      3    4    ×      5    6    ×  ⋯  ×      99    100    =      1    2    ×      2    2    ×      3    4    ×      4    4    ×      5    6    ×      6    6    ×  ⋯  ×      99    100    ×      100    100    A  =      1          4              50                        1      ×      2      ⋯      ×      100              (      1      ×      2      ⋯      ×      50              )        2              =      1          4              50                        100      !              (      50      !              )        2            At this point, we can use Stirling approximation of the factorial  n  !  ≈      2    π    n              (            n    e                      )              n  So      A    n    =            (      2      n      )      !                      4        n            (      n      !              )        2              ≈      1          π      n      For the problem, use   n  =  50

How to show 1 2 </mfrac> ⋅ 3 4 </mfrac> &#

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