Limit of $a_n=(1- frac{1}{n})^n$ as $nrightarrow infty$












1














I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.






calculus







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  • $displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
    – Yadati Kiran
    Nov 25 '18 at 7:30












  • haha, oh wow. Subtle
    – Wesley Strik
    Nov 25 '18 at 7:31










  • It's $-1$ and I am looking at $frac{1}{e}$
    – Wesley Strik
    Nov 25 '18 at 7:31






  • 1




    Thats it! Just look at the general formula.
    – Yadati Kiran
    Nov 25 '18 at 7:32


















1














I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.






calculus







share|cite|improve this question






















  • $displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
    – Yadati Kiran
    Nov 25 '18 at 7:30












  • haha, oh wow. Subtle
    – Wesley Strik
    Nov 25 '18 at 7:31










  • It's $-1$ and I am looking at $frac{1}{e}$
    – Wesley Strik
    Nov 25 '18 at 7:31






  • 1




    Thats it! Just look at the general formula.
    – Yadati Kiran
    Nov 25 '18 at 7:32
















1












1








1







I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.






calculus







share|cite|improve this question













I want to calculate the limit of the following sequence:
$$a_n=(1- frac{1}{n})^n$$
First off I will calculate some terms to understand the behaviour:
$$a_1=0 $$
$$a_2=left(frac{1}{2} right)^2 =frac{1}{4} $$
$$ vdots$$
$$ a_{20}=left(frac{19}{20}right)^{20} approx 0.358$$
$$vdots$$
$$ a_{100}=left(frac{99}{100}right)^{100} approx0.366$$
That seems like a very small number, I would not immediately recognise this as something I am familiar with. So far the exercises I've been doing are all related to $e$ in some way, so maybe I'm simply not recognising where $e$ comes in.






calculus




calculus






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share|cite|improve this question











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asked Nov 25 '18 at 7:28









Wesley Strik

1,493422




1,493422












  • $displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
    – Yadati Kiran
    Nov 25 '18 at 7:30












  • haha, oh wow. Subtle
    – Wesley Strik
    Nov 25 '18 at 7:31










  • It's $-1$ and I am looking at $frac{1}{e}$
    – Wesley Strik
    Nov 25 '18 at 7:31






  • 1




    Thats it! Just look at the general formula.
    – Yadati Kiran
    Nov 25 '18 at 7:32




















  • $displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
    – Yadati Kiran
    Nov 25 '18 at 7:30












  • haha, oh wow. Subtle
    – Wesley Strik
    Nov 25 '18 at 7:31










  • It's $-1$ and I am looking at $frac{1}{e}$
    – Wesley Strik
    Nov 25 '18 at 7:31






  • 1




    Thats it! Just look at the general formula.
    – Yadati Kiran
    Nov 25 '18 at 7:32


















$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30






$displaystyle e^x=lim_{ntoinfty}left(1+dfrac xnright)^n$. What is $x$ in your case here?
– Yadati Kiran
Nov 25 '18 at 7:30














haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31




haha, oh wow. Subtle
– Wesley Strik
Nov 25 '18 at 7:31












It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31




It's $-1$ and I am looking at $frac{1}{e}$
– Wesley Strik
Nov 25 '18 at 7:31




1




1




Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32






Thats it! Just look at the general formula.
– Yadati Kiran
Nov 25 '18 at 7:32












5 Answers
5






active

oldest

votes


















7














Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.



If you are only given $(1 + 1/n)^n to e$, then



$$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$






share|cite|improve this answer





















  • Oh that is even nicer than the argument I came up with just now :D Thank you.
    – Wesley Strik
    Nov 25 '18 at 7:56








  • 1




    @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
    – RRL
    Nov 25 '18 at 8:00



















3














Notice that:
$$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
For very large $n$, we have that by an index shift argument this will have the same limit as:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
$$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$






share|cite|improve this answer





























    1














    Simply write:
    $$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$






    share|cite|improve this answer



















    • 1




      $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
      – Yadati Kiran
      Nov 25 '18 at 7:40





















    1














    You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $






    share|cite|improve this answer





























      1














      Or you could use



      $$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$






      share|cite|improve this answer





















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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7














        Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.



        If you are only given $(1 + 1/n)^n to e$, then



        $$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$






        share|cite|improve this answer





















        • Oh that is even nicer than the argument I came up with just now :D Thank you.
          – Wesley Strik
          Nov 25 '18 at 7:56








        • 1




          @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
          – RRL
          Nov 25 '18 at 8:00
















        7














        Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.



        If you are only given $(1 + 1/n)^n to e$, then



        $$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$






        share|cite|improve this answer





















        • Oh that is even nicer than the argument I came up with just now :D Thank you.
          – Wesley Strik
          Nov 25 '18 at 7:56








        • 1




          @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
          – RRL
          Nov 25 '18 at 8:00














        7












        7








        7






        Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.



        If you are only given $(1 + 1/n)^n to e$, then



        $$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$






        share|cite|improve this answer












        Using $(1 +x/n)^n to e^x$ somewhat defeats the purpose of such a question.



        If you are only given $(1 + 1/n)^n to e$, then



        $$left(1 - frac{1}{n} right)^n = left(frac{n-1}{n} right)^n = frac{1}{left(1 + frac{1}{n-1}right)^{n-1}left(1 + frac{1}{n-1}right)} to frac{1}{e cdot 1} = e^{-1} $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 7:49









        RRL

        48.9k42573




        48.9k42573












        • Oh that is even nicer than the argument I came up with just now :D Thank you.
          – Wesley Strik
          Nov 25 '18 at 7:56








        • 1




          @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
          – RRL
          Nov 25 '18 at 8:00


















        • Oh that is even nicer than the argument I came up with just now :D Thank you.
          – Wesley Strik
          Nov 25 '18 at 7:56








        • 1




          @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
          – RRL
          Nov 25 '18 at 8:00
















        Oh that is even nicer than the argument I came up with just now :D Thank you.
        – Wesley Strik
        Nov 25 '18 at 7:56






        Oh that is even nicer than the argument I came up with just now :D Thank you.
        – Wesley Strik
        Nov 25 '18 at 7:56






        1




        1




        @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
        – RRL
        Nov 25 '18 at 8:00




        @WesleyGroupshaveFeelingsToo: Thanks. You filled in the details nicely.
        – RRL
        Nov 25 '18 at 8:00











        3














        Notice that:
        $$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
        For very large $n$, we have that by an index shift argument this will have the same limit as:
        $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
        Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
        $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$






        share|cite|improve this answer


























          3














          Notice that:
          $$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
          For very large $n$, we have that by an index shift argument this will have the same limit as:
          $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
          Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
          $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$






          share|cite|improve this answer
























            3












            3








            3






            Notice that:
            $$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
            For very large $n$, we have that by an index shift argument this will have the same limit as:
            $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
            Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
            $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$






            share|cite|improve this answer












            Notice that:
            $$a_n=left( frac{n-1}{n}right)^n =left( frac{n}{n-1}right)^{-n}=left( frac{n-1+1}{n-1}right)^{-n}= left(left( 1+frac{1}{n-1}right)^{n} right) ^{-1}$$
            For very large $n$, we have that by an index shift argument this will have the same limit as:
            $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} $$
            Terms with an exponent will have the same limit as the limit of the sequence raised to that power. $(lim_{n rightarrow infty} (a_n )^k = A^k)$ where we denote the limit of $a_n$ by $A$). We thus get:
            $$ left( left( 1+frac{1}{n}right)^{n+1} right)^{-1} rightarrow e^{-1}. $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 '18 at 7:54









            Wesley Strik

            1,493422




            1,493422























                1














                Simply write:
                $$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$






                share|cite|improve this answer



















                • 1




                  $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
                  – Yadati Kiran
                  Nov 25 '18 at 7:40


















                1














                Simply write:
                $$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$






                share|cite|improve this answer



















                • 1




                  $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
                  – Yadati Kiran
                  Nov 25 '18 at 7:40
















                1












                1








                1






                Simply write:
                $$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$






                share|cite|improve this answer














                Simply write:
                $$a_n= left(1 - frac{1}{n} right)^n= left(1 + frac{(-1)}{n} right)^{n } rightarrow e^{-1}=frac{1}{e}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 25 '18 at 7:43

























                answered Nov 25 '18 at 7:37









                Wesley Strik

                1,493422




                1,493422








                • 1




                  $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
                  – Yadati Kiran
                  Nov 25 '18 at 7:40
















                • 1




                  $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
                  – Yadati Kiran
                  Nov 25 '18 at 7:40










                1




                1




                $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
                – Yadati Kiran
                Nov 25 '18 at 7:40






                $left(1 - frac{1}{n} right)^n=left(1 +frac{(-1)}{n} right)^nto e^{-1}$ shall be more appropriate.
                – Yadati Kiran
                Nov 25 '18 at 7:40













                1














                You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $






                share|cite|improve this answer


























                  1














                  You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $






                  share|cite|improve this answer
























                    1












                    1








                    1






                    You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $






                    share|cite|improve this answer












                    You can still write $a_n=left(1-dfrac{1}{n}right)^n$ in terms of $b_n = left(1+dfrac{1}{n}right)^n$. In fact, $a_n = dfrac{b_n}{(b_{frac{n-1}{2}})^{frac{2n}{n-1}}}to dfrac{e}{e^2} = dfrac{1}{e}. $







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 25 '18 at 7:49









                    DeepSea

                    70.9k54487




                    70.9k54487























                        1














                        Or you could use



                        $$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$






                        share|cite|improve this answer


























                          1














                          Or you could use



                          $$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$






                          share|cite|improve this answer
























                            1












                            1








                            1






                            Or you could use



                            $$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$






                            share|cite|improve this answer












                            Or you could use



                            $$lim_{n to infty}bigg(1-frac{1}{n}bigg)^n = lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n(-1)} = bigg[lim_{n to infty}bigg(1+frac{1}{-n}bigg)^{-n}bigg]^{-1} = e^{-1}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 25 '18 at 8:04









                            KM101

                            4,886421




                            4,886421






























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