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Math Help - Infinite Series with Numerator Arithmetic and Denominator Geometric

  1. #1
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    Infinite Series with Numerator Arithmetic and Denominator Geometric

    (1/1) + (4/4) + (7/16) + (10/64) ...

    This series is supposed to be an infinite series but it's not going up by a set common ratio. How do I find the sum of the infinite series of this series?
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  2. #2
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    Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

    Hello, Espionage!

    \text{Evaluate: }\:S \;=\;\frac{1}{1} + \frac{4}{4} + \frac{7}{4^2} + \frac{10}{4^3} + \cdots

    \begin{array}{ccccccc}\text{We are given:}&S &=& \dfrac{1}{1} + \dfrac{4}{4} + \dfrac{7}{4^2} + \dfrac{10}{4^3} + \cdots \\ \\[-3mm] \text{Multiply by }\frac{1}{4}\!: & \frac{1}{4}S &=& \quad\;\;\; \dfrac{1}{4} + \dfrac{4}{4^2} + \dfrac{7}{4^3} + \cdots \\ \\[-3mm] \text{Subtract:} & \frac{3}{4}S &=& 1 + \underbrace{\frac{3}{4} + \frac{3}{4^2} + \frac{3}{4^3} + \cdots}_{\text{geometric series}} \end{array}


    The geometric series has first term a = \tfrac{3}{4} and common ratio r = \tfrac{1}{4}

    Its sum is: . \dfrac{\frac{3}{4}}{1-\frac{1}{4}} \:=\:\dfrac{\frac{3}{4}}{\frac{3}{4}} \:=\:1


    Hence: . \tfrac{3}{4}S \:=\:1+1 \:=\:2

    Therefore: . S \:=\:\frac{8}{3}


    **
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  3. #3
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    Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

    Edit: Post above answered it way better.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

    If I was going to find the infinite sum, I would begin with the difference equation to obtain the partial sum in closed form:

    S_{n}-S_{n-1}=\frac{3n+1}{4^n} where S_0=1

    The corresponding homogeneous solution is:

    h_n=c_1

    We then look for a particular solution of the form:

    p_n=(An+B)4^{-n}

    We may now determine the coefficients A and B.

    Substituting this into the original difference equation, we find:

    (An+B)4^{-n}-(A(n-1)+B)4^{-(n-1)}=(3n+1)4^{-n}

    Multiply through by 4^{n}:

    (An+B)-4(A(n-1)+B)=3n+1

    -3An+4A-3B=3n+1

    Equating coefficients, we obtain the system:

    -3A=3\,\therefore\,A=-1

    4A-3B=1\,\therefore\,B=-\frac{5}{3}

    And thus:

    p_n=-\frac{1}{3}(3n+5)4^{-n}

    By superposition, we have:

    S_n=h_n+p_n=c_1-\frac{1}{3}(3n+5)4^{-n}

    Using the initial value, we find:

    S_0=c_1-\frac{1}{3}(3(0)+5)4^{0}=c_1-\frac{5}{3}=1\,\therefore\,c_1=\frac{8}{3}

    and so we find:

    S_n=\frac{8}{3}-\frac{1}{3}(3n+5)4^{-n}

    Hence, the infinite sum is:

    S=\lim_{n\to\infty}S_n=\frac{8}{3}
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  5. #5
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    Re: Infinite Series with Numerator Arithmetic and Denominator Geometric

    Thank you very much.
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