Summation of series $sum_x=1^inftyfrac-theta^xln(1-theta)$

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How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$



I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.










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    It is a sum of a geometric series.
    – Dr. Sonnhard Graubner
    Sep 8 at 12:03














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down vote

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How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$



I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.










share|cite|improve this question

















  • 2




    It is a sum of a geometric series.
    – Dr. Sonnhard Graubner
    Sep 8 at 12:03












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$



I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.










share|cite|improve this question













How should I simplify $$sum_x=1^inftyfrac-theta^xln(1-theta)$$ into $$frac-theta(1-theta)ln(1-theta)$$



I was doing a statistic question I'm stuck in this step. Hope someone could explain it for me. Thanks in advance.







sequences-and-series






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asked Sep 8 at 12:02









Mathxx

3,37311131




3,37311131







  • 2




    It is a sum of a geometric series.
    – Dr. Sonnhard Graubner
    Sep 8 at 12:03












  • 2




    It is a sum of a geometric series.
    – Dr. Sonnhard Graubner
    Sep 8 at 12:03







2




2




It is a sum of a geometric series.
– Dr. Sonnhard Graubner
Sep 8 at 12:03




It is a sum of a geometric series.
– Dr. Sonnhard Graubner
Sep 8 at 12:03










2 Answers
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Well



$$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$






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    Just because of the formula
    $$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$






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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      up vote
      3
      down vote



      accepted










      Well



      $$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$






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        up vote
        3
        down vote



        accepted










        Well



        $$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$






        share|cite|improve this answer






















          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          Well



          $$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$






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          Well



          $$sum_x=1^inftyfrac-theta^xln(1-theta)=frac-1ln(1-theta)sum_x=1^inftytheta^x=frac-thetaln(1-theta)sum_x=0^inftytheta^x=frac-thetaln(1-theta)cdotfrac11-theta$$







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          answered Sep 8 at 12:08









          b00n heT

          8,90211833




          8,90211833




















              up vote
              2
              down vote













              Just because of the formula
              $$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$






              share|cite|improve this answer
























                up vote
                2
                down vote













                Just because of the formula
                $$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Just because of the formula
                  $$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$






                  share|cite|improve this answer












                  Just because of the formula
                  $$sum_n=k^inftyu^n=u^ksum_n=0^inftyu^n=fracu^k1-u.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 8 at 12:06









                  Bernard

                  112k635104




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