limit of geometric series sum when r = 1












1












$begingroup$


I'm currently learning the prove of sum of geometric series on Khan Academy.



I understand the behaviour of the function when |r| > 1, when |r| < 1, when r = 0 and when r = -1, but I am a bit confused by its behaviour when r = 1.



enter image description here



The narrator said that when r = 1, the limit function is undefined because the denominator of the limit function would be 0, and the behaviour of the limit function is UNDEFINED, which I do understand.



My confusion arises when I tried to substitute r = 1 into the original function for sum of geometric series



enter image description here



if r = 1, then every term would equal to a, and the sum of the geometric series would approach infinity, so its behaviour is DEFINED.



So when r = 1, behaviour of sum function is DEFINED, but behaviour of limit function is UNDEFINED, but sum function also equal to limit function?!



This is causing me so much confusion.






calculus sequences-and-series limits summation







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I'm currently learning the prove of sum of geometric series on Khan Academy.



    I understand the behaviour of the function when |r| > 1, when |r| < 1, when r = 0 and when r = -1, but I am a bit confused by its behaviour when r = 1.



    enter image description here



    The narrator said that when r = 1, the limit function is undefined because the denominator of the limit function would be 0, and the behaviour of the limit function is UNDEFINED, which I do understand.



    My confusion arises when I tried to substitute r = 1 into the original function for sum of geometric series



    enter image description here



    if r = 1, then every term would equal to a, and the sum of the geometric series would approach infinity, so its behaviour is DEFINED.



    So when r = 1, behaviour of sum function is DEFINED, but behaviour of limit function is UNDEFINED, but sum function also equal to limit function?!



    This is causing me so much confusion.






    calculus sequences-and-series limits summation







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I'm currently learning the prove of sum of geometric series on Khan Academy.



      I understand the behaviour of the function when |r| > 1, when |r| < 1, when r = 0 and when r = -1, but I am a bit confused by its behaviour when r = 1.



      enter image description here



      The narrator said that when r = 1, the limit function is undefined because the denominator of the limit function would be 0, and the behaviour of the limit function is UNDEFINED, which I do understand.



      My confusion arises when I tried to substitute r = 1 into the original function for sum of geometric series



      enter image description here



      if r = 1, then every term would equal to a, and the sum of the geometric series would approach infinity, so its behaviour is DEFINED.



      So when r = 1, behaviour of sum function is DEFINED, but behaviour of limit function is UNDEFINED, but sum function also equal to limit function?!



      This is causing me so much confusion.






      calculus sequences-and-series limits summation







      share|cite|improve this question









      $endgroup$




      I'm currently learning the prove of sum of geometric series on Khan Academy.



      I understand the behaviour of the function when |r| > 1, when |r| < 1, when r = 0 and when r = -1, but I am a bit confused by its behaviour when r = 1.



      enter image description here



      The narrator said that when r = 1, the limit function is undefined because the denominator of the limit function would be 0, and the behaviour of the limit function is UNDEFINED, which I do understand.



      My confusion arises when I tried to substitute r = 1 into the original function for sum of geometric series



      enter image description here



      if r = 1, then every term would equal to a, and the sum of the geometric series would approach infinity, so its behaviour is DEFINED.



      So when r = 1, behaviour of sum function is DEFINED, but behaviour of limit function is UNDEFINED, but sum function also equal to limit function?!



      This is causing me so much confusion.






      calculus sequences-and-series limits summation




      calculus sequences-and-series limits summation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 42 mins ago









      ThorThor

      24617




      24617






















          3 Answers
          3






          active

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          3












          $begingroup$

          What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
            $endgroup$
            – Thor
            35 mins ago






          • 1




            $begingroup$
            My first equality holds for every $rneq1$.
            $endgroup$
            – José Carlos Santos
            33 mins ago



















          2












          $begingroup$

          In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with



          $$S_n = 1 + r+ r^2+ ... + r^n$$
          $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$



          Then we subtract both equations.



          $$S_n ( 1-r) = 1 - r^{n+1}$$



          Solving for $S_n$ requires that $rneq 1$ or we would dived by $0$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            +1 for showing why the result fails for $r=1$.
            $endgroup$
            – Paramanand Singh
            23 mins ago



















          0












          $begingroup$

          You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.



          Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.






          share|cite|improve this answer









          $endgroup$













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            3 Answers
            3






            active

            oldest

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            3 Answers
            3






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
              $endgroup$
              – Thor
              35 mins ago






            • 1




              $begingroup$
              My first equality holds for every $rneq1$.
              $endgroup$
              – José Carlos Santos
              33 mins ago
















            3












            $begingroup$

            What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
              $endgroup$
              – Thor
              35 mins ago






            • 1




              $begingroup$
              My first equality holds for every $rneq1$.
              $endgroup$
              – José Carlos Santos
              33 mins ago














            3












            3








            3





            $begingroup$

            What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$






            share|cite|improve this answer









            $endgroup$



            What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 38 mins ago









            José Carlos SantosJosé Carlos Santos

            169k23132237




            169k23132237












            • $begingroup$
              hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
              $endgroup$
              – Thor
              35 mins ago






            • 1




              $begingroup$
              My first equality holds for every $rneq1$.
              $endgroup$
              – José Carlos Santos
              33 mins ago


















            • $begingroup$
              hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
              $endgroup$
              – Thor
              35 mins ago






            • 1




              $begingroup$
              My first equality holds for every $rneq1$.
              $endgroup$
              – José Carlos Santos
              33 mins ago
















            $begingroup$
            hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
            $endgroup$
            – Thor
            35 mins ago




            $begingroup$
            hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you.
            $endgroup$
            – Thor
            35 mins ago




            1




            1




            $begingroup$
            My first equality holds for every $rneq1$.
            $endgroup$
            – José Carlos Santos
            33 mins ago




            $begingroup$
            My first equality holds for every $rneq1$.
            $endgroup$
            – José Carlos Santos
            33 mins ago











            2












            $begingroup$

            In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with



            $$S_n = 1 + r+ r^2+ ... + r^n$$
            $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$



            Then we subtract both equations.



            $$S_n ( 1-r) = 1 - r^{n+1}$$



            Solving for $S_n$ requires that $rneq 1$ or we would dived by $0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for showing why the result fails for $r=1$.
              $endgroup$
              – Paramanand Singh
              23 mins ago
















            2












            $begingroup$

            In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with



            $$S_n = 1 + r+ r^2+ ... + r^n$$
            $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$



            Then we subtract both equations.



            $$S_n ( 1-r) = 1 - r^{n+1}$$



            Solving for $S_n$ requires that $rneq 1$ or we would dived by $0$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              +1 for showing why the result fails for $r=1$.
              $endgroup$
              – Paramanand Singh
              23 mins ago














            2












            2








            2





            $begingroup$

            In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with



            $$S_n = 1 + r+ r^2+ ... + r^n$$
            $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$



            Then we subtract both equations.



            $$S_n ( 1-r) = 1 - r^{n+1}$$



            Solving for $S_n$ requires that $rneq 1$ or we would dived by $0$.






            share|cite|improve this answer









            $endgroup$



            In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with



            $$S_n = 1 + r+ r^2+ ... + r^n$$
            $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$



            Then we subtract both equations.



            $$S_n ( 1-r) = 1 - r^{n+1}$$



            Solving for $S_n$ requires that $rneq 1$ or we would dived by $0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 29 mins ago









            MachineLearnerMachineLearner

            1,071110




            1,071110












            • $begingroup$
              +1 for showing why the result fails for $r=1$.
              $endgroup$
              – Paramanand Singh
              23 mins ago


















            • $begingroup$
              +1 for showing why the result fails for $r=1$.
              $endgroup$
              – Paramanand Singh
              23 mins ago
















            $begingroup$
            +1 for showing why the result fails for $r=1$.
            $endgroup$
            – Paramanand Singh
            23 mins ago




            $begingroup$
            +1 for showing why the result fails for $r=1$.
            $endgroup$
            – Paramanand Singh
            23 mins ago











            0












            $begingroup$

            You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.



            Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.



              Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.



                Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.






                share|cite|improve this answer









                $endgroup$



                You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.



                Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 11 mins ago









                AllawonderAllawonder

                2,256616




                2,256616






























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