How to find limit of a sequence $lim_{nto infty} frac{1+a+a^2+…+a^n}{1+b+b^2+…+b^n}$?












1












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I have no idea how to solve this: $$lim_{nto infty} frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}$$ for: $|a|<1$, $|b|<1 $



I would be happy for any advice.



Thank you






limits







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  • $begingroup$
    Can you show what you have done so far?
    $endgroup$
    – Shubham Johri
    Nov 25 '18 at 21:37










  • $begingroup$
    You should play around with the series: $1+a+a^2 + dots + a^n $. For starters plug in some numbers like a = 2,3 etc. and for n numbers like 2,3,4,5 etc. Maybe you will notice something :)
    $endgroup$
    – Imago
    Nov 25 '18 at 21:39
















1












$begingroup$


I have no idea how to solve this: $$lim_{nto infty} frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}$$ for: $|a|<1$, $|b|<1 $



I would be happy for any advice.



Thank you






limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you show what you have done so far?
    $endgroup$
    – Shubham Johri
    Nov 25 '18 at 21:37










  • $begingroup$
    You should play around with the series: $1+a+a^2 + dots + a^n $. For starters plug in some numbers like a = 2,3 etc. and for n numbers like 2,3,4,5 etc. Maybe you will notice something :)
    $endgroup$
    – Imago
    Nov 25 '18 at 21:39














1












1








1





$begingroup$


I have no idea how to solve this: $$lim_{nto infty} frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}$$ for: $|a|<1$, $|b|<1 $



I would be happy for any advice.



Thank you






limits







share|cite|improve this question









$endgroup$




I have no idea how to solve this: $$lim_{nto infty} frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}$$ for: $|a|<1$, $|b|<1 $



I would be happy for any advice.



Thank you






limits




limits






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 '18 at 21:36









Lauren SinLauren Sin

955




955












  • $begingroup$
    Can you show what you have done so far?
    $endgroup$
    – Shubham Johri
    Nov 25 '18 at 21:37










  • $begingroup$
    You should play around with the series: $1+a+a^2 + dots + a^n $. For starters plug in some numbers like a = 2,3 etc. and for n numbers like 2,3,4,5 etc. Maybe you will notice something :)
    $endgroup$
    – Imago
    Nov 25 '18 at 21:39


















  • $begingroup$
    Can you show what you have done so far?
    $endgroup$
    – Shubham Johri
    Nov 25 '18 at 21:37










  • $begingroup$
    You should play around with the series: $1+a+a^2 + dots + a^n $. For starters plug in some numbers like a = 2,3 etc. and for n numbers like 2,3,4,5 etc. Maybe you will notice something :)
    $endgroup$
    – Imago
    Nov 25 '18 at 21:39
















$begingroup$
Can you show what you have done so far?
$endgroup$
– Shubham Johri
Nov 25 '18 at 21:37




$begingroup$
Can you show what you have done so far?
$endgroup$
– Shubham Johri
Nov 25 '18 at 21:37












$begingroup$
You should play around with the series: $1+a+a^2 + dots + a^n $. For starters plug in some numbers like a = 2,3 etc. and for n numbers like 2,3,4,5 etc. Maybe you will notice something :)
$endgroup$
– Imago
Nov 25 '18 at 21:39




$begingroup$
You should play around with the series: $1+a+a^2 + dots + a^n $. For starters plug in some numbers like a = 2,3 etc. and for n numbers like 2,3,4,5 etc. Maybe you will notice something :)
$endgroup$
– Imago
Nov 25 '18 at 21:39










2 Answers
2






active

oldest

votes


















3












$begingroup$

The limit is $$lim_{nto infty}left(frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}cdotfrac{1-a}{1-b}cdotfrac{1-b}{1-a}right)=lim_{nto infty} frac{1-a^{n+1}}{1-b^{n+1}}cdotfrac{1-b}{1-a}=frac{1-b}{1-a}$$



EDIT



My thinking (pedagogic insert):
$1+a+a^2+...+a^n;$ is the sum of first $n+1$ terms of geometric series.

The calculation that simplifies the sum and helps when looking for the limit, is multiplying by $frac{1-a}{1-a}.$

Similarly for $;1+b+b^2+...+b^n.$

Both numerator and denominator converge (since $|a|,|b|<1$).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
    $endgroup$
    – Lauren Sin
    Nov 26 '18 at 9:30








  • 1




    $begingroup$
    @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
    $endgroup$
    – user376343
    Nov 26 '18 at 12:25





















5












$begingroup$

Hint: $1+x+x^2+ldots + x^n = frac{1-x^{n+1}}{1-x}$.






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









    3












    $begingroup$

    The limit is $$lim_{nto infty}left(frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}cdotfrac{1-a}{1-b}cdotfrac{1-b}{1-a}right)=lim_{nto infty} frac{1-a^{n+1}}{1-b^{n+1}}cdotfrac{1-b}{1-a}=frac{1-b}{1-a}$$



    EDIT



    My thinking (pedagogic insert):
    $1+a+a^2+...+a^n;$ is the sum of first $n+1$ terms of geometric series.

    The calculation that simplifies the sum and helps when looking for the limit, is multiplying by $frac{1-a}{1-a}.$

    Similarly for $;1+b+b^2+...+b^n.$

    Both numerator and denominator converge (since $|a|,|b|<1$).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
      $endgroup$
      – Lauren Sin
      Nov 26 '18 at 9:30








    • 1




      $begingroup$
      @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
      $endgroup$
      – user376343
      Nov 26 '18 at 12:25


















    3












    $begingroup$

    The limit is $$lim_{nto infty}left(frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}cdotfrac{1-a}{1-b}cdotfrac{1-b}{1-a}right)=lim_{nto infty} frac{1-a^{n+1}}{1-b^{n+1}}cdotfrac{1-b}{1-a}=frac{1-b}{1-a}$$



    EDIT



    My thinking (pedagogic insert):
    $1+a+a^2+...+a^n;$ is the sum of first $n+1$ terms of geometric series.

    The calculation that simplifies the sum and helps when looking for the limit, is multiplying by $frac{1-a}{1-a}.$

    Similarly for $;1+b+b^2+...+b^n.$

    Both numerator and denominator converge (since $|a|,|b|<1$).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
      $endgroup$
      – Lauren Sin
      Nov 26 '18 at 9:30








    • 1




      $begingroup$
      @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
      $endgroup$
      – user376343
      Nov 26 '18 at 12:25
















    3












    3








    3





    $begingroup$

    The limit is $$lim_{nto infty}left(frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}cdotfrac{1-a}{1-b}cdotfrac{1-b}{1-a}right)=lim_{nto infty} frac{1-a^{n+1}}{1-b^{n+1}}cdotfrac{1-b}{1-a}=frac{1-b}{1-a}$$



    EDIT



    My thinking (pedagogic insert):
    $1+a+a^2+...+a^n;$ is the sum of first $n+1$ terms of geometric series.

    The calculation that simplifies the sum and helps when looking for the limit, is multiplying by $frac{1-a}{1-a}.$

    Similarly for $;1+b+b^2+...+b^n.$

    Both numerator and denominator converge (since $|a|,|b|<1$).






    share|cite|improve this answer











    $endgroup$



    The limit is $$lim_{nto infty}left(frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}cdotfrac{1-a}{1-b}cdotfrac{1-b}{1-a}right)=lim_{nto infty} frac{1-a^{n+1}}{1-b^{n+1}}cdotfrac{1-b}{1-a}=frac{1-b}{1-a}$$



    EDIT



    My thinking (pedagogic insert):
    $1+a+a^2+...+a^n;$ is the sum of first $n+1$ terms of geometric series.

    The calculation that simplifies the sum and helps when looking for the limit, is multiplying by $frac{1-a}{1-a}.$

    Similarly for $;1+b+b^2+...+b^n.$

    Both numerator and denominator converge (since $|a|,|b|<1$).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 '18 at 18:48

























    answered Nov 25 '18 at 21:48









    user376343user376343

    3,3583826




    3,3583826












    • $begingroup$
      I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
      $endgroup$
      – Lauren Sin
      Nov 26 '18 at 9:30








    • 1




      $begingroup$
      @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
      $endgroup$
      – user376343
      Nov 26 '18 at 12:25




















    • $begingroup$
      I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
      $endgroup$
      – Lauren Sin
      Nov 26 '18 at 9:30








    • 1




      $begingroup$
      @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
      $endgroup$
      – user376343
      Nov 26 '18 at 12:25


















    $begingroup$
    I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
    $endgroup$
    – Lauren Sin
    Nov 26 '18 at 9:30






    $begingroup$
    I have 1 more questions. 1st: Can you please describe what was your way of thinking about this problem? Maybe if you had some more ideas that didnt work, or did you just look at it and it was obvious? Thanks
    $endgroup$
    – Lauren Sin
    Nov 26 '18 at 9:30






    1




    1




    $begingroup$
    @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
    $endgroup$
    – user376343
    Nov 26 '18 at 12:25






    $begingroup$
    @LaurenSin Yes, I did just look, the method was in my memory. To my students I would say whote I completed in the solution.
    $endgroup$
    – user376343
    Nov 26 '18 at 12:25













    5












    $begingroup$

    Hint: $1+x+x^2+ldots + x^n = frac{1-x^{n+1}}{1-x}$.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Hint: $1+x+x^2+ldots + x^n = frac{1-x^{n+1}}{1-x}$.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Hint: $1+x+x^2+ldots + x^n = frac{1-x^{n+1}}{1-x}$.






        share|cite|improve this answer









        $endgroup$



        Hint: $1+x+x^2+ldots + x^n = frac{1-x^{n+1}}{1-x}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 21:40









        EnnarEnnar

        14.4k32443




        14.4k32443






























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