Finding the infinite sum of a geometric series












0












$begingroup$


I have the series:
$3 + frac{3}{2} + frac{3}{4} + frac{3}{8}$,



I need to calculate the exact value of it. I have found that the equation of this is $a_n = 3(frac{1}{2})^{n-1}$, but I am not sure how to progress from here. Would guess that I need to convert it to sigma notation but I am not totally sure.






geometric-series







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  • 2




    $begingroup$
    $$sum_{n=1}^{infty}ax^{n-1} = dfrac{a}{1-x}$$
    $endgroup$
    – David Peterson
    Dec 21 '18 at 4:25
















0












$begingroup$


I have the series:
$3 + frac{3}{2} + frac{3}{4} + frac{3}{8}$,



I need to calculate the exact value of it. I have found that the equation of this is $a_n = 3(frac{1}{2})^{n-1}$, but I am not sure how to progress from here. Would guess that I need to convert it to sigma notation but I am not totally sure.






geometric-series







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $$sum_{n=1}^{infty}ax^{n-1} = dfrac{a}{1-x}$$
    $endgroup$
    – David Peterson
    Dec 21 '18 at 4:25














0












0








0





$begingroup$


I have the series:
$3 + frac{3}{2} + frac{3}{4} + frac{3}{8}$,



I need to calculate the exact value of it. I have found that the equation of this is $a_n = 3(frac{1}{2})^{n-1}$, but I am not sure how to progress from here. Would guess that I need to convert it to sigma notation but I am not totally sure.






geometric-series







share|cite|improve this question









$endgroup$




I have the series:
$3 + frac{3}{2} + frac{3}{4} + frac{3}{8}$,



I need to calculate the exact value of it. I have found that the equation of this is $a_n = 3(frac{1}{2})^{n-1}$, but I am not sure how to progress from here. Would guess that I need to convert it to sigma notation but I am not totally sure.






geometric-series




geometric-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 21 '18 at 4:19









ElijahElijah

626




626








  • 2




    $begingroup$
    $$sum_{n=1}^{infty}ax^{n-1} = dfrac{a}{1-x}$$
    $endgroup$
    – David Peterson
    Dec 21 '18 at 4:25














  • 2




    $begingroup$
    $$sum_{n=1}^{infty}ax^{n-1} = dfrac{a}{1-x}$$
    $endgroup$
    – David Peterson
    Dec 21 '18 at 4:25








2




2




$begingroup$
$$sum_{n=1}^{infty}ax^{n-1} = dfrac{a}{1-x}$$
$endgroup$
– David Peterson
Dec 21 '18 at 4:25




$begingroup$
$$sum_{n=1}^{infty}ax^{n-1} = dfrac{a}{1-x}$$
$endgroup$
– David Peterson
Dec 21 '18 at 4:25










2 Answers
2






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oldest

votes


















2












$begingroup$

You have the terms right: you're sum is $$sum_{n=1}^infty 3left(frac{1}{2}right)^{n-1}=3sum_{n=0}^inftyleft(frac{1}{2}right)^n$$
Now recall the geometric series formula is $$sum_{n=0}^infty x^n=frac{1}{1-x}qquad |x|<1$$ So what is $x$ in your case?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    $x = frac{1}{2}$ which would make the answer $2$?
    $endgroup$
    – Elijah
    Dec 21 '18 at 4:27










  • $begingroup$
    Yes, but don't forget you're multiplying the geometric series by $3$.
    $endgroup$
    – Dave
    Dec 21 '18 at 4:28






  • 1




    $begingroup$
    Ohhh, making the final answer 6? Thanks for the help!
    $endgroup$
    – Elijah
    Dec 21 '18 at 4:30



















0












$begingroup$

I'm going to assume this is supposed to go on forever. First, I'm going to solve this problem, then show the general case.



Suppose $S = 3 + frac{3}{2} + frac{3}{4} + cdots$. Note that $frac{S}{2} = frac{3}{2} + frac{3}{4} + frac{3}{8} + cdots$ Now, we note that $S - frac{S}{2} = 3$, because the second sequence is missing nothing but the first term. We can then solve to show that $S = boxed{6}$.



Now, note we have the general geometric sequence $S = a + a times x + a times x^2 + cdots$. We know that $S times x = a times x + a times x^2 + a times x^3 + cdots rightarrow S - S times x = a rightarrow S = boxed{frac{a}{1-x}}$. Note this only works for $|x| < 1$, or else the geometric sequence would not converge.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    2












    $begingroup$

    You have the terms right: you're sum is $$sum_{n=1}^infty 3left(frac{1}{2}right)^{n-1}=3sum_{n=0}^inftyleft(frac{1}{2}right)^n$$
    Now recall the geometric series formula is $$sum_{n=0}^infty x^n=frac{1}{1-x}qquad |x|<1$$ So what is $x$ in your case?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $x = frac{1}{2}$ which would make the answer $2$?
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:27










    • $begingroup$
      Yes, but don't forget you're multiplying the geometric series by $3$.
      $endgroup$
      – Dave
      Dec 21 '18 at 4:28






    • 1




      $begingroup$
      Ohhh, making the final answer 6? Thanks for the help!
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:30
















    2












    $begingroup$

    You have the terms right: you're sum is $$sum_{n=1}^infty 3left(frac{1}{2}right)^{n-1}=3sum_{n=0}^inftyleft(frac{1}{2}right)^n$$
    Now recall the geometric series formula is $$sum_{n=0}^infty x^n=frac{1}{1-x}qquad |x|<1$$ So what is $x$ in your case?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      $x = frac{1}{2}$ which would make the answer $2$?
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:27










    • $begingroup$
      Yes, but don't forget you're multiplying the geometric series by $3$.
      $endgroup$
      – Dave
      Dec 21 '18 at 4:28






    • 1




      $begingroup$
      Ohhh, making the final answer 6? Thanks for the help!
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:30














    2












    2








    2





    $begingroup$

    You have the terms right: you're sum is $$sum_{n=1}^infty 3left(frac{1}{2}right)^{n-1}=3sum_{n=0}^inftyleft(frac{1}{2}right)^n$$
    Now recall the geometric series formula is $$sum_{n=0}^infty x^n=frac{1}{1-x}qquad |x|<1$$ So what is $x$ in your case?






    share|cite|improve this answer









    $endgroup$



    You have the terms right: you're sum is $$sum_{n=1}^infty 3left(frac{1}{2}right)^{n-1}=3sum_{n=0}^inftyleft(frac{1}{2}right)^n$$
    Now recall the geometric series formula is $$sum_{n=0}^infty x^n=frac{1}{1-x}qquad |x|<1$$ So what is $x$ in your case?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 4:25









    DaveDave

    9,08711033




    9,08711033












    • $begingroup$
      $x = frac{1}{2}$ which would make the answer $2$?
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:27










    • $begingroup$
      Yes, but don't forget you're multiplying the geometric series by $3$.
      $endgroup$
      – Dave
      Dec 21 '18 at 4:28






    • 1




      $begingroup$
      Ohhh, making the final answer 6? Thanks for the help!
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:30


















    • $begingroup$
      $x = frac{1}{2}$ which would make the answer $2$?
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:27










    • $begingroup$
      Yes, but don't forget you're multiplying the geometric series by $3$.
      $endgroup$
      – Dave
      Dec 21 '18 at 4:28






    • 1




      $begingroup$
      Ohhh, making the final answer 6? Thanks for the help!
      $endgroup$
      – Elijah
      Dec 21 '18 at 4:30
















    $begingroup$
    $x = frac{1}{2}$ which would make the answer $2$?
    $endgroup$
    – Elijah
    Dec 21 '18 at 4:27




    $begingroup$
    $x = frac{1}{2}$ which would make the answer $2$?
    $endgroup$
    – Elijah
    Dec 21 '18 at 4:27












    $begingroup$
    Yes, but don't forget you're multiplying the geometric series by $3$.
    $endgroup$
    – Dave
    Dec 21 '18 at 4:28




    $begingroup$
    Yes, but don't forget you're multiplying the geometric series by $3$.
    $endgroup$
    – Dave
    Dec 21 '18 at 4:28




    1




    1




    $begingroup$
    Ohhh, making the final answer 6? Thanks for the help!
    $endgroup$
    – Elijah
    Dec 21 '18 at 4:30




    $begingroup$
    Ohhh, making the final answer 6? Thanks for the help!
    $endgroup$
    – Elijah
    Dec 21 '18 at 4:30











    0












    $begingroup$

    I'm going to assume this is supposed to go on forever. First, I'm going to solve this problem, then show the general case.



    Suppose $S = 3 + frac{3}{2} + frac{3}{4} + cdots$. Note that $frac{S}{2} = frac{3}{2} + frac{3}{4} + frac{3}{8} + cdots$ Now, we note that $S - frac{S}{2} = 3$, because the second sequence is missing nothing but the first term. We can then solve to show that $S = boxed{6}$.



    Now, note we have the general geometric sequence $S = a + a times x + a times x^2 + cdots$. We know that $S times x = a times x + a times x^2 + a times x^3 + cdots rightarrow S - S times x = a rightarrow S = boxed{frac{a}{1-x}}$. Note this only works for $|x| < 1$, or else the geometric sequence would not converge.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I'm going to assume this is supposed to go on forever. First, I'm going to solve this problem, then show the general case.



      Suppose $S = 3 + frac{3}{2} + frac{3}{4} + cdots$. Note that $frac{S}{2} = frac{3}{2} + frac{3}{4} + frac{3}{8} + cdots$ Now, we note that $S - frac{S}{2} = 3$, because the second sequence is missing nothing but the first term. We can then solve to show that $S = boxed{6}$.



      Now, note we have the general geometric sequence $S = a + a times x + a times x^2 + cdots$. We know that $S times x = a times x + a times x^2 + a times x^3 + cdots rightarrow S - S times x = a rightarrow S = boxed{frac{a}{1-x}}$. Note this only works for $|x| < 1$, or else the geometric sequence would not converge.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I'm going to assume this is supposed to go on forever. First, I'm going to solve this problem, then show the general case.



        Suppose $S = 3 + frac{3}{2} + frac{3}{4} + cdots$. Note that $frac{S}{2} = frac{3}{2} + frac{3}{4} + frac{3}{8} + cdots$ Now, we note that $S - frac{S}{2} = 3$, because the second sequence is missing nothing but the first term. We can then solve to show that $S = boxed{6}$.



        Now, note we have the general geometric sequence $S = a + a times x + a times x^2 + cdots$. We know that $S times x = a times x + a times x^2 + a times x^3 + cdots rightarrow S - S times x = a rightarrow S = boxed{frac{a}{1-x}}$. Note this only works for $|x| < 1$, or else the geometric sequence would not converge.






        share|cite|improve this answer









        $endgroup$



        I'm going to assume this is supposed to go on forever. First, I'm going to solve this problem, then show the general case.



        Suppose $S = 3 + frac{3}{2} + frac{3}{4} + cdots$. Note that $frac{S}{2} = frac{3}{2} + frac{3}{4} + frac{3}{8} + cdots$ Now, we note that $S - frac{S}{2} = 3$, because the second sequence is missing nothing but the first term. We can then solve to show that $S = boxed{6}$.



        Now, note we have the general geometric sequence $S = a + a times x + a times x^2 + cdots$. We know that $S times x = a times x + a times x^2 + a times x^3 + cdots rightarrow S - S times x = a rightarrow S = boxed{frac{a}{1-x}}$. Note this only works for $|x| < 1$, or else the geometric sequence would not converge.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 4:27









        Neeyanth KopparapuNeeyanth Kopparapu

        3016




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