How prove this H 2 n − H n + 1 4 n &gt; ln ⁡ 2 Show that, for every positive integer n, 1 n + 1 + 1 n + 2 + ⋯ + 1 2 n + 1 4 n &gt; ln ⁡ 2 I know this lim n → ∞ 1 n + 1 + 1 n + 2 + ⋯ + 1 2 n = ln ⁡ 2and use this ln ⁡ ( 1 + 1 n ) &lt; 1 n is not usefulBut for this inequality I can't. Thank you

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How prove this       H          2      n        −      H          n        +      1          4      n        &amp;gt;  ln  ⁡      2  Show that, for every positive integer   n,            1              n        +        1              +            1              n        +        2              +  ⋯  +            1              2        n              +            1              4        n              &amp;gt;  ln  ⁡      2  I know this      lim          n      →      ∞                  1              n        +        1              +            1              n        +        2              +  ⋯  +            1              2        n              =  ln  ⁡  2and use this  ln  ⁡      (    1    +                  1        n              )    &amp;lt;            1      n      is not usefulBut for this inequality I can&#039;t. Thank you

floygdarvn · Accepted Answer

A not at all tedious computation shows that the expression - which I rearrange a tiny bit -                              1                      2            n                          +                  1                      n            +            1                          +        …        +                  1                      2            n            −            1                          +                  1                      4            n                                              =                  ∑                      k            =            0                                n            −            1                                    1          2                          (                      1                          n              +              k                                +                      1                          n              +              k              +              1                                )                                                  =                  1          n                          ∑                      k            =            0                                n            −            1                                    1          2                          (                      1                          1              +                              k                n                                              +                      1                          1              +                                                k                  +                  1                                n                                              )                    is a trapezium sum for the integral      ∫    0    1              d      t              1      +      t        =  log  ⁡  2  ,and since       1          1      +      t       is convex, the trapeziums have greater area than the corresponding part of the integral.

skottyrottenmf · Accepted Answer

Let       A    n    =      H          2      n        −      H          n        +      1          4      n      , then a tedious but easy computation shows that      A          n      +      1        −      A    n    =            −      1              4      n      (      n      +      1      )      (      2      n      +      1      )        &amp;lt;  0  ,hence   (      A    n    ) is decreasing, in particular,       A    n    &amp;gt;  A where   A  =      lim          n      →      ∞            A    n  . Since       H    n    =  log  ⁡  n  +  γ  +  o  (  1  ) when   n  →  ∞,   A  =  log  ⁡  2. QED.Remark: The funny thing here (and the subject of the next questions of the exercise as it is usually asked) is that the same technique applies to the sequence       B    n    =      H          2      n        −      H          n      , showing that   (      B    n    ) is increasing, hence       B    n    &amp;lt;  B where   B  =      lim          n      →      ∞            B    n  . Since   B  =  log  ⁡  2 as well, this yields some approximations of   log  ⁡  2, namely the fact that, for every   n  ⩾  1,      H          2      n        −      H          n        &amp;lt;  log  ⁡  2  &amp;lt;      H          2      n        −      H          n        +      1          4      n        .Likewise, for every   n  ⩾  1,      H          2      n        −      H          n        +      1          4      n      +      2        &amp;lt;  log  ⁡  2  &amp;lt;      H          2      n        −      H          n        +      1          4      n        ,thus, the correction       1          4      n       is asymptotically of the right order.

How prove this H <mrow class="MJX-TeXAtom-ORD"> 2 n </mrow> </m

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