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Evaluate the limit the following series


Solve limit with Riemann sum.Rearranging Limit ProblemLimit of a function with absolute valuesEvaluate the following limit without L'Hopitalevaluate the limit $lim_xto 2+frac[x]sin (x-2)(x-2)^2$Calculate the sum of a seriesDetermine whether the series $frace^frac1nn^2$ converges or divergesA limit involving irrational powers: can I use L'Hopital?Compute the limit as $t to 0$ of the given expressionLimit of a Sequence involving factorialsHow do you evaluate $lim_xtoinfty(x!*e^-x^2)$






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$



Evaluate the limit $$lim_nto infty sum_j=1^n frac4j^2n^3.$$




I was under the assumption that this would just tend to $0$ after expanding everything because the denominator will grow quicker than the numerator, but apparently this is not the case.






sequences-and-series limits limits-without-lhopital







share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    The keyword is "Riemann sum". See for example math.stackexchange.com/questions/2712723/…?
    $endgroup$
    – Robert Z
    8 hours ago


















2












$begingroup$



Evaluate the limit $$lim_nto infty sum_j=1^n frac4j^2n^3.$$




I was under the assumption that this would just tend to $0$ after expanding everything because the denominator will grow quicker than the numerator, but apparently this is not the case.






sequences-and-series limits limits-without-lhopital







share|cite|improve this question











$endgroup$







  • 3




    $begingroup$
    The keyword is "Riemann sum". See for example math.stackexchange.com/questions/2712723/…?
    $endgroup$
    – Robert Z
    8 hours ago














2












2








2





$begingroup$



Evaluate the limit $$lim_nto infty sum_j=1^n frac4j^2n^3.$$




I was under the assumption that this would just tend to $0$ after expanding everything because the denominator will grow quicker than the numerator, but apparently this is not the case.






sequences-and-series limits limits-without-lhopital







share|cite|improve this question











$endgroup$





Evaluate the limit $$lim_nto infty sum_j=1^n frac4j^2n^3.$$




I was under the assumption that this would just tend to $0$ after expanding everything because the denominator will grow quicker than the numerator, but apparently this is not the case.






sequences-and-series limits limits-without-lhopital




sequences-and-series limits limits-without-lhopital






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









Bach

2,5652 gold badges9 silver badges24 bronze badges




2,5652 gold badges9 silver badges24 bronze badges










asked 8 hours ago









WallaceWallace

1587 bronze badges




1587 bronze badges







  • 3




    $begingroup$
    The keyword is "Riemann sum". See for example math.stackexchange.com/questions/2712723/…?
    $endgroup$
    – Robert Z
    8 hours ago













  • 3




    $begingroup$
    The keyword is "Riemann sum". See for example math.stackexchange.com/questions/2712723/…?
    $endgroup$
    – Robert Z
    8 hours ago








3




3




$begingroup$
The keyword is "Riemann sum". See for example math.stackexchange.com/questions/2712723/…?
$endgroup$
– Robert Z
8 hours ago





$begingroup$
The keyword is "Riemann sum". See for example math.stackexchange.com/questions/2712723/…?
$endgroup$
– Robert Z
8 hours ago











2 Answers
2






active

oldest

votes


















4












$begingroup$

Recognize the limit as the integral $$int _0^1 4x^2 dx $$






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    Note that $$sum_j=1^n j^2=fracn(n+1)(2n+1)6$$ we have
    beginalign
    lim_ntoinftysum_j=1^nfrac4j^2n^3&=lim_ntoinftyfrac2(n+1)(2n+1)3n^2\
    &=frac23lim_ntoinfty(1+frac1n)(2+frac 1n)\
    &=frac43.
    endalign






    share|cite|improve this answer









    $endgroup$















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      2 Answers
      2






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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Recognize the limit as the integral $$int _0^1 4x^2 dx $$






      share|cite|improve this answer









      $endgroup$

















        4












        $begingroup$

        Recognize the limit as the integral $$int _0^1 4x^2 dx $$






        share|cite|improve this answer









        $endgroup$















          4












          4








          4





          $begingroup$

          Recognize the limit as the integral $$int _0^1 4x^2 dx $$






          share|cite|improve this answer









          $endgroup$



          Recognize the limit as the integral $$int _0^1 4x^2 dx $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Mohammad Riazi-KermaniMohammad Riazi-Kermani

          48.9k4 gold badges27 silver badges71 bronze badges




          48.9k4 gold badges27 silver badges71 bronze badges























              2












              $begingroup$

              Note that $$sum_j=1^n j^2=fracn(n+1)(2n+1)6$$ we have
              beginalign
              lim_ntoinftysum_j=1^nfrac4j^2n^3&=lim_ntoinftyfrac2(n+1)(2n+1)3n^2\
              &=frac23lim_ntoinfty(1+frac1n)(2+frac 1n)\
              &=frac43.
              endalign






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                Note that $$sum_j=1^n j^2=fracn(n+1)(2n+1)6$$ we have
                beginalign
                lim_ntoinftysum_j=1^nfrac4j^2n^3&=lim_ntoinftyfrac2(n+1)(2n+1)3n^2\
                &=frac23lim_ntoinfty(1+frac1n)(2+frac 1n)\
                &=frac43.
                endalign






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Note that $$sum_j=1^n j^2=fracn(n+1)(2n+1)6$$ we have
                  beginalign
                  lim_ntoinftysum_j=1^nfrac4j^2n^3&=lim_ntoinftyfrac2(n+1)(2n+1)3n^2\
                  &=frac23lim_ntoinfty(1+frac1n)(2+frac 1n)\
                  &=frac43.
                  endalign






                  share|cite|improve this answer









                  $endgroup$



                  Note that $$sum_j=1^n j^2=fracn(n+1)(2n+1)6$$ we have
                  beginalign
                  lim_ntoinftysum_j=1^nfrac4j^2n^3&=lim_ntoinftyfrac2(n+1)(2n+1)3n^2\
                  &=frac23lim_ntoinfty(1+frac1n)(2+frac 1n)\
                  &=frac43.
                  endalign







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 8 hours ago









                  BachBach

                  2,5652 gold badges9 silver badges24 bronze badges




                  2,5652 gold badges9 silver badges24 bronze badges



























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