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Math Help - Sum to infinity of unusual geometric series

  1. #1
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    Sum to infinity of unusual geometric series

    There is a problem that says: the series S(k) is a geometric series with S(2) = 425 and S(4) = 25. Find the different possible values for S(infinity).

    So far I have seen that the common ratio is an imaginary number, and I think it is i (4/sqrt(17)), and I have to see what is the sum to infinity of the series by writing down some terms. There seems to be a connexion with Euler's series, but I really don't know what to do. Could somebody help?
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  2. #2
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    Since neither term is 0 we can say that a \neq 0 (this will come in handy later when it comes to cancelling a which can only be done for non-zero values)

    ar = 425 and ar^3 = 25

    \dfrac{ar^3}{ar} = \dfrac{25}{425} \implies r^2 = \dfrac{1}{17}

    Since \dfrac{1}{17} > 0 there will be two real solutions for r which can then be plugged into the standard formula. I've no idea how you get r to be complex
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    Quote Originally Posted by e^(i*pi) View Post
    Since neither term is 0 we can say that a \neq 0 (this will come in handy later when it comes to cancelling a which can only be done for non-zero values)

    ar = 425 and ar^3 = 25

    \dfrac{ar^3}{ar} = \dfrac{25}{425} \implies r^2 = \dfrac{1}{17}

    Since \dfrac{1}{17} > 0 there will be two real solutions for r which can then be plugged into the standard formula. I've no idea how you get r to be complex
    Yes, I understand what you mean, but look, they don't give you any term in the sequence, they only give you the sum of the series of the terms, that its, 425 is not ar, but a+ar, and 25 is a + ar + ar^2 + ar3.
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    Quote Originally Posted by Kurama View Post
    There is a problem that says: the series S(k) is a geometric series with S(2) = 425 and S(4) = 25. Find the different possible values for S(infinity).

    So far I have seen that the common ratio is an imaginary number, and I think it is i (4/sqrt(17)), and I have to see what is the sum to infinity of the series by writing down some terms. There seems to be a connexion with Euler's series, but I really don't know what to do. Could somebody help?


    \displaystyle{25=S(4)=a_1\,\frac{q^4-1}{q-1}\,,\,\,425=S(2)=a_1\,\frac{q^2-1}{q-1}\Longrightarrow \frac{1}{17}=\frac{25}{425}=q^2+1\Longrightarrow

    \displaystyle{ q^2=-\frac{16}{17}\Longrightarrow q=\pm \frac{4}{\sqrt{17}}\,i} , and you are right.

    Now, I can't understand how come you have to evaluate the infinite series by "writing down

    some terms"...why would you want to do that? Use the formula!

    \displaystyle{q=\pm \frac{4}{\sqrt{17}}\Longrightarrow 425=S(2)=a_1\,\frac{q^2-1}{q-1}=a_1(q+1)\Longrightarrow a_1=\frac{425}{q+1}=\alpha , so:

    \displaystyle{S(\infty)=\frac{\alpha}{1-q} , and we're done.

    Note that there are two different values for \alpha ...

    Tonio
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    Thank you for the help. Now, I suppose that we could substitute the expression for alpha in the final result to obtain
    425 / (q2 - 1), and I'd say that this means that both values of alpha give the same result, right? The reason why I was trying to write down some terms is that I know that the formula can be used when q<1, but I had no idea that it could also be used for complex numbers with modulus less than 1, so I was trying to see what happened when you raised q to infinity.
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  6. #6
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    Quote Originally Posted by Kurama View Post
    Thank you for the help. Now, I suppose that we could substitute the expression for alpha in the final result to obtain
    425 / (q2 - 1), and I'd say that this means that both values of alpha give the same result, right? The reason why I was trying to write down some terms is that I know that the formula can be used when q<1, but I had no idea that it could also be used for complex numbers with modulus less than 1, so I was trying to see what happened when you raised q to infinity.


    Of course you can use the formula with complex number of modulus less than 1, but I didn't understand

    why did you deduce that both values of alpha give the same result: no, they don't, and the

    formula for S(\infty) in my last post makes this clear, imo.

    Tonio
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