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Math Help - Compute the following summation please(!)

  1. #1
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    Compute the following summation please(!)

    The sum (from n=1 to infinity) of (n^2)(1/3)^n

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  2. #2
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    Hello, feyomi!

    \text{Evaluate: }\;S \;=\;\sum^{\infty}_{n=1} \frac{n^2}{3^n}

    \begin{array}{cccccccccc}<br />
\text{We have:} & S &=& \dfrac{1}{3} + \dfrac{4}{3^2} + \dfrac{9}{3^3} + \dfrac{16}{3^4} + \dfrac{25}{3^5} + \dfrac{36}{3^6} + \hdots \\ \\[-2mm]<br />
\text{Multiply by }\frac{1}{3}\!: & \dfrac{1}{3}S &=& \quad\;\;  \dfrac{1}{3^2} + \dfrac{4}{3^3} + \dfrac{9}{3^4} + \dfrac{16}{3^5} + \dfrac{25}{3^6} + \hdots \\\end{array}

    \begin{array}{ccccccc}\text{Subtract: } & \qquad \dfrac{2}{3}S &=& \dfrac{1}{3} + \dfrac{3}{3^2} + \dfrac{5}{3^3} + \dfrac{7}{3^4} + \dfrac{9}{3^5} + \dfrac{11}{3^6} + \hdots   \end{array}

    \begin{array}{cccccc}\text{Multiply by }\frac{1}{3}\!: & \dfrac{2}{9}S &=& \qquad \dfrac{1}{3^2} + \dfrac{3}{3^3} + \dfrac{5}{3^4} + \dfrac{7}{3^5} + \dfrac{9}{3^6} = \hdots \\<br />
\end{array}

    \begin{array}{ccccccc}\text{Subtract:} & \qquad\;\dfrac{4}{9}S &=& \dfrac{1}{3} + \dfrac{2}{3^2} + \dfrac{2}{3^3} + \dfrac{2}{3^4} + \dfrac{2}{3^5} + \dfrac{2}{3^6} + \hdots  \end{array}


    \text{We have: }\;\frac{4}{9}S \;=\;\frac{1}{3} + \frac{2}{3}\underbrace{\left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \hdots \right)}_{\text{geometric series}}

    . . The geometric series has the sum: . \frac{1}{1-\frac{1}{3}} \:=\:\frac{3}{2}


    Hence: . \frac{4}{9}S \;=\;\frac{1}{3} + \frac{2}{3}\left(\frac{3}{2}\right) \;=\;\frac{1}{3} + 1 \quad\Rightarrow\quad \frac{4}{9}S\;=\;\frac{4}{3}


    Therefore: . S \;=\;\frac{9}{4}\cdot\frac{4}{3} \quad\Rightarrow\quad \boxed{S\;=\;3}

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  3. #3
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    I think that another possible method for this problem would be summation by parts (I did not try myself), but Soroban's approach is very nice.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by feyomi View Post
    The sum (from n=1 to infinity) of (n^2)(1/3)^n

    Thanks
    The other classical maneuver is to notice that \sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in  fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    The other classical maneuver is to notice that \sum_{n=1}^{\infty}n^2x^n=x^2\left(\sum_{n=0}^{\in  fty}x^n\right)''=x^2\left(\frac{1}{1-x}\right)''
    I think that sort of argument requires uniform convergence so that you can swap the limit (there is an implicit limit when you take the derivative) unless you choose to treat things as formal power series, though then the question of convergence can't really be answered.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by gmatt View Post
    I think that sort of argument requires uniform convergence so that you can swap the limit (there is an implicit limit when you take the derivative) unless you choose to treat things as formal power series, though then the question of convergence can't really be answered.
    power series converge unilformly on their disk of convergence
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    power series converge unilformly on their disk of convergence
    Yeap thats right, just pointing out the trick is being used implicitly.
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