Limit of a harmonic subseries minus "its" logarithm lim n → ∞ ∑ k = 1 n 1 3 k − 1 − 1 3 ln ⁡ ( n ) I think that inserting the other terms and then subtracting them would not help. I need just the ideea. Thank you.

Question

Limit of a harmonic subseries minus &quot;its&quot; logarithm            lim              n        →        ∞                    ∑              k        =        1            n              1              3        k        −        1              −          1      3        ln    ⁡    (    n    )  I think that inserting the other terms and then subtracting them would not help. I need just the ideea. Thank you.

Quinn Everett · Accepted Answer

It is not very difficult to show that, for natural values of a,                     ∑                  k          =          1                n                    1              n        +        a              =          H              n        +        a              −          H      a       , where       H    n   is the       n          t      h       harmonic number. But in our case,   a  =  −      1    3  ,  so we have to extend the formula or definition of       H    n   so as to encompass non-natural arguments as well. Luckily for us, Euler has already done it some   250 years ago:                     H        n            =              ∫        0        1                              1          −                      x            n                                    1          −          x                    d      x       . Using this, we can now compute the value of       H          −              1        3             to be                     1        2                              (                            π                  3                    −      3      ln      ⁡      3                        )                    . Now use the fact that                     lim                  n          →          ∞                            H        n            −      ln      ⁡      n        =    γ  . See here for more details on how to handle the integral in question.

pokoljitef2 · Accepted Answer

If it were       ∑          k      =      1        n        1    k    −  ln  ⁡  n , we&#039;d end up with   γ. So I&#039;d suggest to investigate      ∑          k      =      1        ∞        (          3              3        k        −        1              −          1      k        )

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