Let γ n = ( 1 + 1 2 + . . . 1 n ) − ∫ 1 n 1 t d t . Show that the sequence { γ n } 1 n converges.

Question

Let       γ    n    =  (  1  +      1    2    +  .  .  .      1    n    )  −      ∫    1    n        1    t    d  t  .   Show that the sequence   {      γ    n        }    1    n     converges.

Adeline Shah · Accepted Answer

Since   f  :  [  1  ,  ∞  )  →      R   such that   f  (  x  )  =  1      /    x  , for all   x  ≥  1  , is a decreasing function, we have            1              n        +        1              ≤            1      x        ≤            1      n        ,for all   n  ≤  x  ≤  n  +  1. Hence,      ∫          n              n      +      1                  1      t        d  t  ≤            1      n        .Thus,      ∑          i      =      1              k            ∫          i              i      +      1                  1      t        d  t  ≤      ∑          i      =      1              k                  1      i        ,which implies that  0  ≤      ∑          i      =      1              k                  1      i        −      ∑          i      =      1              k            ∫          i              i      +      1                  1      t        d  t  =      ∑          i      =      1              k                  1      i        −      ∫          1              k      +      1                  1      t        d  t  ≤      ∑          i      =      1              k                  1      i        −      ∫          1              k                  1      t        d  t  =      γ          k        .This means that   {      γ    k    } is a decreasing and bounded below by zero sequence, meaning it converges.

Let γ n </msub> = ( 1 + 1 2 </mfrac> + . .

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