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Limit of geometric series sum when $r = 1$


Testing if a geometric series converges by taking limit to infinityElementary converge and divergeVertical asymptote of $frac 3x^2 - 18x - 816x^2 - 54$Intuition for gradient when you only have one variable?Find the limit $limlimits_n to inftyfrac2^-n^2sumlimits_k=n+1^infty 2^-k^2$Sum of a finite geometric seriesShowing that the sum of a complex power series is equal to an indicator variableWhy this limit exists (Khan Academy)Sum of Geometric Series FormulaCan we say a function is “unbounded” when we mean it''s tending to infinity?













3












$begingroup$


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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    "proof" ≠ "prove".
    $endgroup$
    – user21820
    yesterday










  • $begingroup$
    I would argue both the series and the rational function give the value $infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner.
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $mathbf R.$ At best, it has both $+infty$ and $-infty$ as candidate values.
    $endgroup$
    – Allawonder
    yesterday










  • $begingroup$
    @Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $mathbb R$
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-infty.$
    $endgroup$
    – Allawonder
    yesterday















3












$begingroup$


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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    "proof" ≠ "prove".
    $endgroup$
    – user21820
    yesterday










  • $begingroup$
    I would argue both the series and the rational function give the value $infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner.
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $mathbf R.$ At best, it has both $+infty$ and $-infty$ as candidate values.
    $endgroup$
    – Allawonder
    yesterday










  • $begingroup$
    @Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $mathbb R$
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-infty.$
    $endgroup$
    – Allawonder
    yesterday













3












3








3





$begingroup$


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.










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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Benjamin

625419




625419










asked yesterday









ThorThor

26117




26117











  • $begingroup$
    "proof" ≠ "prove".
    $endgroup$
    – user21820
    yesterday










  • $begingroup$
    I would argue both the series and the rational function give the value $infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner.
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $mathbf R.$ At best, it has both $+infty$ and $-infty$ as candidate values.
    $endgroup$
    – Allawonder
    yesterday










  • $begingroup$
    @Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $mathbb R$
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-infty.$
    $endgroup$
    – Allawonder
    yesterday
















  • $begingroup$
    "proof" ≠ "prove".
    $endgroup$
    – user21820
    yesterday










  • $begingroup$
    I would argue both the series and the rational function give the value $infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner.
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $mathbf R.$ At best, it has both $+infty$ and $-infty$ as candidate values.
    $endgroup$
    – Allawonder
    yesterday










  • $begingroup$
    @Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $mathbb R$
    $endgroup$
    – Brevan Ellefsen
    yesterday










  • $begingroup$
    @BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-infty.$
    $endgroup$
    – Allawonder
    yesterday















$begingroup$
"proof" ≠ "prove".
$endgroup$
– user21820
yesterday




$begingroup$
"proof" ≠ "prove".
$endgroup$
– user21820
yesterday












$begingroup$
I would argue both the series and the rational function give the value $infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner.
$endgroup$
– Brevan Ellefsen
yesterday




$begingroup$
I would argue both the series and the rational function give the value $infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner.
$endgroup$
– Brevan Ellefsen
yesterday












$begingroup$
@BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $mathbf R.$ At best, it has both $+infty$ and $-infty$ as candidate values.
$endgroup$
– Allawonder
yesterday




$begingroup$
@BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $mathbf R.$ At best, it has both $+infty$ and $-infty$ as candidate values.
$endgroup$
– Allawonder
yesterday












$begingroup$
@Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $mathbb R$
$endgroup$
– Brevan Ellefsen
yesterday




$begingroup$
@Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $mathbb R$
$endgroup$
– Brevan Ellefsen
yesterday












$begingroup$
@BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-infty.$
$endgroup$
– Allawonder
yesterday




$begingroup$
@BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-infty.$
$endgroup$
– Allawonder
yesterday










5 Answers
5






active

oldest

votes


















10












$begingroup$

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
    yesterday


















7












$begingroup$

What happens is that the equality$$sum_k=0^nar^n=fraca-ar^n+11-r$$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
    yesterday






  • 1




    $begingroup$
    My first equality holds for every $rneq1$.
    $endgroup$
    – José Carlos Santos
    yesterday


















2












$begingroup$

It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.



But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    The best way is to look at an actual geometric series with ratio of 1, such as



    $2+2+2+2+2+2+2...$



    Here, because each term is simply the previous term multiplied by 1, the series diverges, no limit can be found for obvious reasons.



    Take the common ratio of $-1$



    $(1)+(-1)+(1)+(-1)+(1)...$



    Here, the value bounces between 0 and 1, so no limit can be found.



    Note: A limit ONLY occurs if the partial sum, i.e. the sum up to n terms, tends to some number as n gets larger and larger.



    As this is not the case if $|r|=1$, the function simply has no limit.






    share|cite|improve this answer








    New contributor




    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$




















      1












      $begingroup$

      The only problem is that you are considering that infinite is a number. If something is infinite, then it's undefined. So, when r = 1, it's not possible to calcule its sum, because the generated infinite value it's not defined.






      share|cite|improve this answer









      $endgroup$












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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        10












        $begingroup$

        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
          yesterday















        10












        $begingroup$

        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
          yesterday













        10












        10








        10





        $begingroup$

        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 yesterday









        MachineLearnerMachineLearner

        1,317111




        1,317111











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
















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















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




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











        7












        $begingroup$

        What happens is that the equality$$sum_k=0^nar^n=fraca-ar^n+11-r$$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
          yesterday






        • 1




          $begingroup$
          My first equality holds for every $rneq1$.
          $endgroup$
          – José Carlos Santos
          yesterday















        7












        $begingroup$

        What happens is that the equality$$sum_k=0^nar^n=fraca-ar^n+11-r$$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
          yesterday






        • 1




          $begingroup$
          My first equality holds for every $rneq1$.
          $endgroup$
          – José Carlos Santos
          yesterday













        7












        7








        7





        $begingroup$

        What happens is that the equality$$sum_k=0^nar^n=fraca-ar^n+11-r$$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$$sum_k=0^nar^n=fraca-ar^n+11-r$$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 yesterday









        José Carlos SantosJosé Carlos Santos

        170k23132238




        170k23132238











        • $begingroup$
          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
          yesterday






        • 1




          $begingroup$
          My first equality holds for every $rneq1$.
          $endgroup$
          – José Carlos Santos
          yesterday
















        • $begingroup$
          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
          yesterday






        • 1




          $begingroup$
          My first equality holds for every $rneq1$.
          $endgroup$
          – José Carlos Santos
          yesterday















        $begingroup$
        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
        yesterday




        $begingroup$
        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
        yesterday




        1




        1




        $begingroup$
        My first equality holds for every $rneq1$.
        $endgroup$
        – José Carlos Santos
        yesterday




        $begingroup$
        My first equality holds for every $rneq1$.
        $endgroup$
        – José Carlos Santos
        yesterday











        2












        $begingroup$

        It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.



        But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.






        share|cite|improve this answer









        $endgroup$

















          2












          $begingroup$

          It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.



          But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.






          share|cite|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$

            It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.



            But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.






            share|cite|improve this answer









            $endgroup$



            It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.



            But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            Yves DaoustYves Daoust

            131k676229




            131k676229





















                1












                $begingroup$

                The best way is to look at an actual geometric series with ratio of 1, such as



                $2+2+2+2+2+2+2...$



                Here, because each term is simply the previous term multiplied by 1, the series diverges, no limit can be found for obvious reasons.



                Take the common ratio of $-1$



                $(1)+(-1)+(1)+(-1)+(1)...$



                Here, the value bounces between 0 and 1, so no limit can be found.



                Note: A limit ONLY occurs if the partial sum, i.e. the sum up to n terms, tends to some number as n gets larger and larger.



                As this is not the case if $|r|=1$, the function simply has no limit.






                share|cite|improve this answer








                New contributor




                aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$

















                  1












                  $begingroup$

                  The best way is to look at an actual geometric series with ratio of 1, such as



                  $2+2+2+2+2+2+2...$



                  Here, because each term is simply the previous term multiplied by 1, the series diverges, no limit can be found for obvious reasons.



                  Take the common ratio of $-1$



                  $(1)+(-1)+(1)+(-1)+(1)...$



                  Here, the value bounces between 0 and 1, so no limit can be found.



                  Note: A limit ONLY occurs if the partial sum, i.e. the sum up to n terms, tends to some number as n gets larger and larger.



                  As this is not the case if $|r|=1$, the function simply has no limit.






                  share|cite|improve this answer








                  New contributor




                  aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    The best way is to look at an actual geometric series with ratio of 1, such as



                    $2+2+2+2+2+2+2...$



                    Here, because each term is simply the previous term multiplied by 1, the series diverges, no limit can be found for obvious reasons.



                    Take the common ratio of $-1$



                    $(1)+(-1)+(1)+(-1)+(1)...$



                    Here, the value bounces between 0 and 1, so no limit can be found.



                    Note: A limit ONLY occurs if the partial sum, i.e. the sum up to n terms, tends to some number as n gets larger and larger.



                    As this is not the case if $|r|=1$, the function simply has no limit.






                    share|cite|improve this answer








                    New contributor




                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    The best way is to look at an actual geometric series with ratio of 1, such as



                    $2+2+2+2+2+2+2...$



                    Here, because each term is simply the previous term multiplied by 1, the series diverges, no limit can be found for obvious reasons.



                    Take the common ratio of $-1$



                    $(1)+(-1)+(1)+(-1)+(1)...$



                    Here, the value bounces between 0 and 1, so no limit can be found.



                    Note: A limit ONLY occurs if the partial sum, i.e. the sum up to n terms, tends to some number as n gets larger and larger.



                    As this is not the case if $|r|=1$, the function simply has no limit.







                    share|cite|improve this answer








                    New contributor




                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered yesterday









                    amanaman

                    926




                    926




                    New contributor




                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





                    New contributor





                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    aman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.





















                        1












                        $begingroup$

                        The only problem is that you are considering that infinite is a number. If something is infinite, then it's undefined. So, when r = 1, it's not possible to calcule its sum, because the generated infinite value it's not defined.






                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          The only problem is that you are considering that infinite is a number. If something is infinite, then it's undefined. So, when r = 1, it's not possible to calcule its sum, because the generated infinite value it's not defined.






                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            The only problem is that you are considering that infinite is a number. If something is infinite, then it's undefined. So, when r = 1, it's not possible to calcule its sum, because the generated infinite value it's not defined.






                            share|cite|improve this answer









                            $endgroup$



                            The only problem is that you are considering that infinite is a number. If something is infinite, then it's undefined. So, when r = 1, it's not possible to calcule its sum, because the generated infinite value it's not defined.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            shewlongshewlong

                            212




                            212



























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