Determine the value ..

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Taken from W.J Kaczor Books Problem in mathematical analysis page no ;28



determine the value



$$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3$$



My attempt :$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=sum_i=1^nfrac1n^3 sum_j=1^ij$



After that im not able to proceed further



pliz help me, Any hints/solution will be appreciated



thanks u






real-analysis algebra-precalculus limits







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    up vote
    2
    down vote

    favorite
    1












    Taken from W.J Kaczor Books Problem in mathematical analysis page no ;28



    determine the value



    $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3$$



    My attempt :$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=sum_i=1^nfrac1n^3 sum_j=1^ij$



    After that im not able to proceed further



    pliz help me, Any hints/solution will be appreciated



    thanks u






    real-analysis algebra-precalculus limits







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      Taken from W.J Kaczor Books Problem in mathematical analysis page no ;28



      determine the value



      $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3$$



      My attempt :$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=sum_i=1^nfrac1n^3 sum_j=1^ij$



      After that im not able to proceed further



      pliz help me, Any hints/solution will be appreciated



      thanks u






      real-analysis algebra-precalculus limits







      share|cite|improve this question















      Taken from W.J Kaczor Books Problem in mathematical analysis page no ;28



      determine the value



      $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3$$



      My attempt :$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=sum_i=1^nfrac1n^3 sum_j=1^ij$



      After that im not able to proceed further



      pliz help me, Any hints/solution will be appreciated



      thanks u






      real-analysis algebra-precalculus limits




      real-analysis algebra-precalculus limits






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      edited 1 hour ago









      Servaes

      19.3k33686




      19.3k33686










      asked 1 hour ago









      jasmine

      1,127213




      1,127213




















          2 Answers
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          accepted










          Assuming is a positive integer, the well known formulas
          $$sum_k=1^mk=fracm(m+1)2
          qquadtext and qquad
          sum_k=1^mk^2=fracm(m+1)(2m+1)6,$$
          yield with some basic algebra
          begineqnarray*
          sum_i=1^nsum_j=1^i fracjn^3
          &=&sum_i=1^nfrac1n^3sum_j=1^ij=frac1n^3sum_i=1^nfraci(i+1)2\
          &=&frac12n^3sum_i=1^ni+frac12n^3sum_i=1^ni^2\
          &=&frac12n^3fracn(n+1)2+frac12n^3fracn(n+1)(2n+1)6\
          &=&frac2n^3+6n^2+3n12n^3=frac16+frac12n+frac13n^2.
          endeqnarray*
          Hence the limit is
          $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=lim_ntoinftyleft(frac16+frac12n+frac13n^2right)=frac16.$$






          share|cite|improve this answer






















          • thanks a lots @servaes
            – jasmine
            1 hour ago

















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          down vote













          Use the formula for $sum_j=1^i j$.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            Assuming is a positive integer, the well known formulas
            $$sum_k=1^mk=fracm(m+1)2
            qquadtext and qquad
            sum_k=1^mk^2=fracm(m+1)(2m+1)6,$$
            yield with some basic algebra
            begineqnarray*
            sum_i=1^nsum_j=1^i fracjn^3
            &=&sum_i=1^nfrac1n^3sum_j=1^ij=frac1n^3sum_i=1^nfraci(i+1)2\
            &=&frac12n^3sum_i=1^ni+frac12n^3sum_i=1^ni^2\
            &=&frac12n^3fracn(n+1)2+frac12n^3fracn(n+1)(2n+1)6\
            &=&frac2n^3+6n^2+3n12n^3=frac16+frac12n+frac13n^2.
            endeqnarray*
            Hence the limit is
            $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=lim_ntoinftyleft(frac16+frac12n+frac13n^2right)=frac16.$$






            share|cite|improve this answer






















            • thanks a lots @servaes
              – jasmine
              1 hour ago














            up vote
            2
            down vote



            accepted










            Assuming is a positive integer, the well known formulas
            $$sum_k=1^mk=fracm(m+1)2
            qquadtext and qquad
            sum_k=1^mk^2=fracm(m+1)(2m+1)6,$$
            yield with some basic algebra
            begineqnarray*
            sum_i=1^nsum_j=1^i fracjn^3
            &=&sum_i=1^nfrac1n^3sum_j=1^ij=frac1n^3sum_i=1^nfraci(i+1)2\
            &=&frac12n^3sum_i=1^ni+frac12n^3sum_i=1^ni^2\
            &=&frac12n^3fracn(n+1)2+frac12n^3fracn(n+1)(2n+1)6\
            &=&frac2n^3+6n^2+3n12n^3=frac16+frac12n+frac13n^2.
            endeqnarray*
            Hence the limit is
            $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=lim_ntoinftyleft(frac16+frac12n+frac13n^2right)=frac16.$$






            share|cite|improve this answer






















            • thanks a lots @servaes
              – jasmine
              1 hour ago












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Assuming is a positive integer, the well known formulas
            $$sum_k=1^mk=fracm(m+1)2
            qquadtext and qquad
            sum_k=1^mk^2=fracm(m+1)(2m+1)6,$$
            yield with some basic algebra
            begineqnarray*
            sum_i=1^nsum_j=1^i fracjn^3
            &=&sum_i=1^nfrac1n^3sum_j=1^ij=frac1n^3sum_i=1^nfraci(i+1)2\
            &=&frac12n^3sum_i=1^ni+frac12n^3sum_i=1^ni^2\
            &=&frac12n^3fracn(n+1)2+frac12n^3fracn(n+1)(2n+1)6\
            &=&frac2n^3+6n^2+3n12n^3=frac16+frac12n+frac13n^2.
            endeqnarray*
            Hence the limit is
            $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=lim_ntoinftyleft(frac16+frac12n+frac13n^2right)=frac16.$$






            share|cite|improve this answer














            Assuming is a positive integer, the well known formulas
            $$sum_k=1^mk=fracm(m+1)2
            qquadtext and qquad
            sum_k=1^mk^2=fracm(m+1)(2m+1)6,$$
            yield with some basic algebra
            begineqnarray*
            sum_i=1^nsum_j=1^i fracjn^3
            &=&sum_i=1^nfrac1n^3sum_j=1^ij=frac1n^3sum_i=1^nfraci(i+1)2\
            &=&frac12n^3sum_i=1^ni+frac12n^3sum_i=1^ni^2\
            &=&frac12n^3fracn(n+1)2+frac12n^3fracn(n+1)(2n+1)6\
            &=&frac2n^3+6n^2+3n12n^3=frac16+frac12n+frac13n^2.
            endeqnarray*
            Hence the limit is
            $$lim_n rightarrow infty sum_i=1^nsum_j=1^i fracjn^3=lim_ntoinftyleft(frac16+frac12n+frac13n^2right)=frac16.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            Servaes

            19.3k33686




            19.3k33686











            • thanks a lots @servaes
              – jasmine
              1 hour ago
















            • thanks a lots @servaes
              – jasmine
              1 hour ago















            thanks a lots @servaes
            – jasmine
            1 hour ago




            thanks a lots @servaes
            – jasmine
            1 hour ago










            up vote
            2
            down vote













            Use the formula for $sum_j=1^i j$.






            share|cite|improve this answer
























              up vote
              2
              down vote













              Use the formula for $sum_j=1^i j$.






              share|cite|improve this answer






















                up vote
                2
                down vote










                up vote
                2
                down vote









                Use the formula for $sum_j=1^i j$.






                share|cite|improve this answer












                Use the formula for $sum_j=1^i j$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Stockfish

                30615




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