Sum of series with binary parity in the numeratorI'm now stuck with this question, and I don't even know where to start: Find sum of series ∑ 1 ∞ f ( n ) n ( n + 1 ) , where f(n) - number of ones in binary representation of n.I wish I could post some moves, that I've tried but I don't know what to do.Thanks!

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Sum of series with binary parity in the numeratorI&#039;m now stuck with this question, and I don&#039;t even know where to start: Find sum of series      ∑    1    ∞              f      (      n      )              n      (      n      +      1      )      , where f(n) - number of ones in binary representation of n.I wish I could post some moves, that I&#039;ve tried but I don&#039;t know what to do.Thanks!

iceniessyoy · Accepted Answer

Maybe there is a quicker way, but here is one. Let the sum be S. We have            f      (      n      )              n      (      n      +      1      )        =            f      (      n      )        n    −            f      (      n      )              n      +      1          ⟹              f      (      2      n      )              (      2      n      )      (      2      n      +      1      )        +            f      (      2      n      +      1      )              (      2      n      +      1      )      (      2      n      +      2      )        =            f      (      2      n      )              2      n        −            f      (      2      n      )      −      f      (      2      n      +      1      )              2      n      +      1        −            f      (      2      n      +      1      )              2      n      +      2      Now   f  (  2  n  +  1  )  =  f  (  2  n  )  +  1  ,    f  (  2  n  )  =  f  (  n  ), so we can write:            f      (      2      n      )              (      2      n      )      (      2      n      +      1      )        +            f      (      2      n      +      1      )              (      2      n      +      1      )      (      2      n      +      2      )        =            f      (      2      n      )              2      n        +      1          2      n      +      1        −            f      (      2      n      )      +      1              2      n      +      2          =      1    2        (                  f        (        n        )            n        −                  f        (        n        )                    n        +        1              )    +      (          1              2        n        +        1              −          1              2        n        +        2              )      ⟹    S  =      1    2    +      1    2        ∑          n      =      1        ∞              f      (      n      )              n      (      n      +      1      )        +      ∑          n      =      1        ∞        (          1              2        n        +        1              −          1              2        n        +        2              )      ⟹    2  S  =  1  +  S  +  2  log  ⁡  2  −  1    ⟹    S  =  2  log  ⁡  2  ≈  1.386

polivijuye · Accepted Answer

Code:double sum=0;for(int i=1;i&amp;lt;9999999;i++){    double s2=Integer.bitCount(i);    sum+=(s2)/(i*(i+1));}Output Data                    n                    ∑                            9                    1.065079365079365                            99                    1.3394382621894894                            999                    1.3800972409478014                            9999                    1.3854974129587205                            99999                    1.3852676077956714                                  (        limit of data type, therefore decreased value        )

Sum of series with binary parity in the numerator I'm now stuck with this question, and I don't eve

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