How to find the partial sum of $n/2^n$? [duplicate]

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  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I'm trying to find a formula for the partial sum of $n/2^n$.



I've tried this so far...



$$ S_n = frac12 + frac24 + frac38 + cdots + fracn2^n $$



Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $frac2^nncdotfracn+12^n+1$ (a terms that should take each term in the sequence to the next term.)



$$ fracn+12n S_n = frac24 + frac38 + cdots + fracn+12^n+1 $$



Then



$$ S_n- fracn+12n S_n = frac12 - fracn+12^n+1$$



$$ S_n = fracfrac12 - fracn+12^n+11-fracn+12n $$



But, alas, this does not give the correct answers.



Can someone point out where I went wrong? Thanks




sequences-and-series




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marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, max_zorn, Parcly Taxel Jul 26 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    I think this might be useful.
    – ippiki-ookami
    Jul 25 at 7:36






  • 1




    $dfracn+12n S_n neq dfrac24 + dfrac38 + cdots + dfracn+12^n+1$ in your starting step. $left( dfrac12 times dfracn+12n neq dfrac24 right)$.
    – Rahul Goswami
    Jul 25 at 7:42











  • Hmmm, I see. I thought because that was all expanded, somehow the rules were different. :) Guess not
    – Ryan Stull
    Jul 25 at 7:46














up vote
1
down vote

favorite













This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I'm trying to find a formula for the partial sum of $n/2^n$.



I've tried this so far...



$$ S_n = frac12 + frac24 + frac38 + cdots + fracn2^n $$



Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $frac2^nncdotfracn+12^n+1$ (a terms that should take each term in the sequence to the next term.)



$$ fracn+12n S_n = frac24 + frac38 + cdots + fracn+12^n+1 $$



Then



$$ S_n- fracn+12n S_n = frac12 - fracn+12^n+1$$



$$ S_n = fracfrac12 - fracn+12^n+11-fracn+12n $$



But, alas, this does not give the correct answers.



Can someone point out where I went wrong? Thanks




sequences-and-series




share|cite|improve this question











marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, max_zorn, Parcly Taxel Jul 26 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    I think this might be useful.
    – ippiki-ookami
    Jul 25 at 7:36






  • 1




    $dfracn+12n S_n neq dfrac24 + dfrac38 + cdots + dfracn+12^n+1$ in your starting step. $left( dfrac12 times dfracn+12n neq dfrac24 right)$.
    – Rahul Goswami
    Jul 25 at 7:42











  • Hmmm, I see. I thought because that was all expanded, somehow the rules were different. :) Guess not
    – Ryan Stull
    Jul 25 at 7:46












up vote
1
down vote

favorite









up vote
1
down vote

favorite












This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I'm trying to find a formula for the partial sum of $n/2^n$.



I've tried this so far...



$$ S_n = frac12 + frac24 + frac38 + cdots + fracn2^n $$



Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $frac2^nncdotfracn+12^n+1$ (a terms that should take each term in the sequence to the next term.)



$$ fracn+12n S_n = frac24 + frac38 + cdots + fracn+12^n+1 $$



Then



$$ S_n- fracn+12n S_n = frac12 - fracn+12^n+1$$



$$ S_n = fracfrac12 - fracn+12^n+11-fracn+12n $$



But, alas, this does not give the correct answers.



Can someone point out where I went wrong? Thanks




sequences-and-series




share|cite|improve this question












This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers



I'm trying to find a formula for the partial sum of $n/2^n$.



I've tried this so far...



$$ S_n = frac12 + frac24 + frac38 + cdots + fracn2^n $$



Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $frac2^nncdotfracn+12^n+1$ (a terms that should take each term in the sequence to the next term.)



$$ fracn+12n S_n = frac24 + frac38 + cdots + fracn+12^n+1 $$



Then



$$ S_n- fracn+12n S_n = frac12 - fracn+12^n+1$$



$$ S_n = fracfrac12 - fracn+12^n+11-fracn+12n $$



But, alas, this does not give the correct answers.



Can someone point out where I went wrong? Thanks





This question already has an answer here:



  • Formula for calculating $sum_n=0^mnr^n$

    4 answers





sequences-and-series





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 25 at 7:32









Ryan Stull

237215




237215




marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, max_zorn, Parcly Taxel Jul 26 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Hans Lundmark, Adrian Keister, José Carlos Santos, max_zorn, Parcly Taxel Jul 26 at 1:09


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









  • 1




    I think this might be useful.
    – ippiki-ookami
    Jul 25 at 7:36






  • 1




    $dfracn+12n S_n neq dfrac24 + dfrac38 + cdots + dfracn+12^n+1$ in your starting step. $left( dfrac12 times dfracn+12n neq dfrac24 right)$.
    – Rahul Goswami
    Jul 25 at 7:42











  • Hmmm, I see. I thought because that was all expanded, somehow the rules were different. :) Guess not
    – Ryan Stull
    Jul 25 at 7:46












  • 1




    I think this might be useful.
    – ippiki-ookami
    Jul 25 at 7:36






  • 1




    $dfracn+12n S_n neq dfrac24 + dfrac38 + cdots + dfracn+12^n+1$ in your starting step. $left( dfrac12 times dfracn+12n neq dfrac24 right)$.
    – Rahul Goswami
    Jul 25 at 7:42











  • Hmmm, I see. I thought because that was all expanded, somehow the rules were different. :) Guess not
    – Ryan Stull
    Jul 25 at 7:46







1




1




I think this might be useful.
– ippiki-ookami
Jul 25 at 7:36




I think this might be useful.
– ippiki-ookami
Jul 25 at 7:36




1




1




$dfracn+12n S_n neq dfrac24 + dfrac38 + cdots + dfracn+12^n+1$ in your starting step. $left( dfrac12 times dfracn+12n neq dfrac24 right)$.
– Rahul Goswami
Jul 25 at 7:42





$dfracn+12n S_n neq dfrac24 + dfrac38 + cdots + dfracn+12^n+1$ in your starting step. $left( dfrac12 times dfracn+12n neq dfrac24 right)$.
– Rahul Goswami
Jul 25 at 7:42













Hmmm, I see. I thought because that was all expanded, somehow the rules were different. :) Guess not
– Ryan Stull
Jul 25 at 7:46




Hmmm, I see. I thought because that was all expanded, somehow the rules were different. :) Guess not
– Ryan Stull
Jul 25 at 7:46










3 Answers
3






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up vote
3
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Just multiply by $frac12$, not that more complicated factor.
$$frac12S_n=frac14+frac28+frac316+dots+frac n2^n+1$$
$$S_n-frac12S_n=frac12+frac14+frac18+dots+frac12^n-frac n2^n+1$$
$$frac12S_n=1-frac12^n-frac n2^n+1$$
$$S_n=2left(1-frac12^n-frac n2^n+1right)$$






share|cite|improve this answer




























    up vote
    2
    down vote













    Observe



    $$2S_n = 1+ frac1+12 + frac2+14 + cdots + frac(n-1)+12^n-1
    \=S_n-1+1+frac12+frac14cdots+frac12^n-1
    \=S_n-frac n2^n+2-frac12^n-1.$$






    share|cite|improve this answer






























      up vote
      1
      down vote













      Hint



      Consider
      $$S_n=sum_k=1^n k x^k=xsum_k=1^n k x^k-1=x left(sum_k=1^n x^k right)'$$



      Just finish and make $x=frac 12$.






      share|cite|improve this answer






























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        3
        down vote



        accepted










        Just multiply by $frac12$, not that more complicated factor.
        $$frac12S_n=frac14+frac28+frac316+dots+frac n2^n+1$$
        $$S_n-frac12S_n=frac12+frac14+frac18+dots+frac12^n-frac n2^n+1$$
        $$frac12S_n=1-frac12^n-frac n2^n+1$$
        $$S_n=2left(1-frac12^n-frac n2^n+1right)$$






        share|cite|improve this answer

























          up vote
          3
          down vote



          accepted










          Just multiply by $frac12$, not that more complicated factor.
          $$frac12S_n=frac14+frac28+frac316+dots+frac n2^n+1$$
          $$S_n-frac12S_n=frac12+frac14+frac18+dots+frac12^n-frac n2^n+1$$
          $$frac12S_n=1-frac12^n-frac n2^n+1$$
          $$S_n=2left(1-frac12^n-frac n2^n+1right)$$






          share|cite|improve this answer























            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            Just multiply by $frac12$, not that more complicated factor.
            $$frac12S_n=frac14+frac28+frac316+dots+frac n2^n+1$$
            $$S_n-frac12S_n=frac12+frac14+frac18+dots+frac12^n-frac n2^n+1$$
            $$frac12S_n=1-frac12^n-frac n2^n+1$$
            $$S_n=2left(1-frac12^n-frac n2^n+1right)$$






            share|cite|improve this answer













            Just multiply by $frac12$, not that more complicated factor.
            $$frac12S_n=frac14+frac28+frac316+dots+frac n2^n+1$$
            $$S_n-frac12S_n=frac12+frac14+frac18+dots+frac12^n-frac n2^n+1$$
            $$frac12S_n=1-frac12^n-frac n2^n+1$$
            $$S_n=2left(1-frac12^n-frac n2^n+1right)$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered Jul 25 at 7:39









            Parcly Taxel

            33.5k136588




            33.5k136588




















                up vote
                2
                down vote













                Observe



                $$2S_n = 1+ frac1+12 + frac2+14 + cdots + frac(n-1)+12^n-1
                \=S_n-1+1+frac12+frac14cdots+frac12^n-1
                \=S_n-frac n2^n+2-frac12^n-1.$$






                share|cite|improve this answer



























                  up vote
                  2
                  down vote













                  Observe



                  $$2S_n = 1+ frac1+12 + frac2+14 + cdots + frac(n-1)+12^n-1
                  \=S_n-1+1+frac12+frac14cdots+frac12^n-1
                  \=S_n-frac n2^n+2-frac12^n-1.$$






                  share|cite|improve this answer

























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









                    Observe



                    $$2S_n = 1+ frac1+12 + frac2+14 + cdots + frac(n-1)+12^n-1
                    \=S_n-1+1+frac12+frac14cdots+frac12^n-1
                    \=S_n-frac n2^n+2-frac12^n-1.$$






                    share|cite|improve this answer















                    Observe



                    $$2S_n = 1+ frac1+12 + frac2+14 + cdots + frac(n-1)+12^n-1
                    \=S_n-1+1+frac12+frac14cdots+frac12^n-1
                    \=S_n-frac n2^n+2-frac12^n-1.$$







                    share|cite|improve this answer















                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jul 25 at 8:06


























                    answered Jul 25 at 7:39









                    Yves Daoust

                    111k665203




                    111k665203




















                        up vote
                        1
                        down vote













                        Hint



                        Consider
                        $$S_n=sum_k=1^n k x^k=xsum_k=1^n k x^k-1=x left(sum_k=1^n x^k right)'$$



                        Just finish and make $x=frac 12$.






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          Hint



                          Consider
                          $$S_n=sum_k=1^n k x^k=xsum_k=1^n k x^k-1=x left(sum_k=1^n x^k right)'$$



                          Just finish and make $x=frac 12$.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            Hint



                            Consider
                            $$S_n=sum_k=1^n k x^k=xsum_k=1^n k x^k-1=x left(sum_k=1^n x^k right)'$$



                            Just finish and make $x=frac 12$.






                            share|cite|improve this answer















                            Hint



                            Consider
                            $$S_n=sum_k=1^n k x^k=xsum_k=1^n k x^k-1=x left(sum_k=1^n x^k right)'$$



                            Just finish and make $x=frac 12$.







                            share|cite|improve this answer















                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jul 25 at 8:11


























                            answered Jul 25 at 7:39









                            Claude Leibovici

                            111k1055126




                            111k1055126












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