Series - $sum_{i=1}^infty (frac{5}{12})^i$ - geometric series?












2












$begingroup$


I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion







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  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57


















2












$begingroup$


I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57
















2












2








2





$begingroup$


I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion







share|cite|improve this question











$endgroup$




I have to solve - $$sum_{i=1}^infty left(frac{5}{12}right)^i$$ - geometric series?



The geometric series sequence I know is - $$sum_{i=0}^infty x_i= frac{1}{1-x}$$



However in my assignment, the series starts from $i=1$.



The solution I have is - $$sum_{i=1}^infty left(frac{5}{12}right)^i = frac{1}{1-frac{5}{12}}-1$$



Can you explain please why is that the solution?






calculus sequences-and-series taylor-expansion




calculus sequences-and-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 14:04









Larry

2,41331129




2,41331129










asked Jan 16 at 13:57









AlanAlan

1,3841021




1,3841021








  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57
















  • 4




    $begingroup$
    Hint: Write out the first few terms.
    $endgroup$
    – Botond
    Jan 16 at 13:57










4




4




$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57






$begingroup$
Hint: Write out the first few terms.
$endgroup$
– Botond
Jan 16 at 13:57












3 Answers
3






active

oldest

votes


















5












$begingroup$

HINT:
$$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






share|cite|improve this answer









$endgroup$









  • 3




    $begingroup$
    which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
    $endgroup$
    – Alan
    Jan 16 at 14:01










  • $begingroup$
    exactly correct @Alan
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:01










  • $begingroup$
    Thank you for your answer and for your time.
    $endgroup$
    – Alan
    Jan 16 at 14:02










  • $begingroup$
    no problem @Alan .. you can mark the answer as correct if you found it useful :)
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:03






  • 1




    $begingroup$
    Yes, in a few minutes when I can :-)
    $endgroup$
    – Alan
    Jan 16 at 14:03



















1












$begingroup$

There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
(setting $;ileftarrow i-1$), so
$$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, a great point of view.
    $endgroup$
    – Alan
    Jan 16 at 14:23



















1












$begingroup$

Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03
















    5












    $begingroup$

    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03














    5












    5








    5





    $begingroup$

    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$






    share|cite|improve this answer









    $endgroup$



    HINT:
    $$sum_{i=0}^infty x_i= frac{1}{1-x} =x_0 + sum_{i=1}^infty x_i$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 13:59









    Ahmad BazziAhmad Bazzi

    8,3282824




    8,3282824








    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03














    • 3




      $begingroup$
      which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
      $endgroup$
      – Alan
      Jan 16 at 14:01










    • $begingroup$
      exactly correct @Alan
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:01










    • $begingroup$
      Thank you for your answer and for your time.
      $endgroup$
      – Alan
      Jan 16 at 14:02










    • $begingroup$
      no problem @Alan .. you can mark the answer as correct if you found it useful :)
      $endgroup$
      – Ahmad Bazzi
      Jan 16 at 14:03






    • 1




      $begingroup$
      Yes, in a few minutes when I can :-)
      $endgroup$
      – Alan
      Jan 16 at 14:03








    3




    3




    $begingroup$
    which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
    $endgroup$
    – Alan
    Jan 16 at 14:01




    $begingroup$
    which means $frac{5}{12}^0 = 1$, plus the rest of the series. Great, thanks.
    $endgroup$
    – Alan
    Jan 16 at 14:01












    $begingroup$
    exactly correct @Alan
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:01




    $begingroup$
    exactly correct @Alan
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:01












    $begingroup$
    Thank you for your answer and for your time.
    $endgroup$
    – Alan
    Jan 16 at 14:02




    $begingroup$
    Thank you for your answer and for your time.
    $endgroup$
    – Alan
    Jan 16 at 14:02












    $begingroup$
    no problem @Alan .. you can mark the answer as correct if you found it useful :)
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:03




    $begingroup$
    no problem @Alan .. you can mark the answer as correct if you found it useful :)
    $endgroup$
    – Ahmad Bazzi
    Jan 16 at 14:03




    1




    1




    $begingroup$
    Yes, in a few minutes when I can :-)
    $endgroup$
    – Alan
    Jan 16 at 14:03




    $begingroup$
    Yes, in a few minutes when I can :-)
    $endgroup$
    – Alan
    Jan 16 at 14:03











    1












    $begingroup$

    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23
















    1












    $begingroup$

    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23














    1












    1








    1





    $begingroup$

    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$






    share|cite|improve this answer









    $endgroup$



    There is another, faster, explanation, directly linked to the nature of the series: you can factor out $x$ from each term of the series:
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=xsum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i-1}= x sum_{i=0}^infty left(frac{5}{12}right)^{mkern-5mu i}$$
    (setting $;ileftarrow i-1$), so
    $$sum_{i=1}^infty left(frac{5}{12}right)^{mkern-5mu i}=frac x{1-x}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 16 at 14:20









    BernardBernard

    121k740116




    121k740116












    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23


















    • $begingroup$
      Thank you, a great point of view.
      $endgroup$
      – Alan
      Jan 16 at 14:23
















    $begingroup$
    Thank you, a great point of view.
    $endgroup$
    – Alan
    Jan 16 at 14:23




    $begingroup$
    Thank you, a great point of view.
    $endgroup$
    – Alan
    Jan 16 at 14:23











    1












    $begingroup$

    Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.






        share|cite|improve this answer









        $endgroup$



        Observe that for suitable $x$: $$(1-x)sum_{i=k}^{infty}x^i=sum_{i=k}^{infty}x^i-sum_{i=k+1}^{infty}x^i=x^k$$so that:$$sum_{i=k}^{infty}x^i=frac{x^k}{1-x}$$It remains to substitute $k=1$ and $x=frac5{12}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 16 at 14:40









        drhabdrhab

        102k545136




        102k545136






























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