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Infinite Series' Summation

  1. May 26, 2013 #1

    AGNuke

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    Given S, an Infinite Series Summation, find [itex]\frac{1728}{485}S[/itex]
    [tex]S=1^2+\frac{3^2}{5^2}+\frac{5^2}{5^4}+\frac{7^2}{5^6}+...[/tex]
    I found out the formula for (r+1)th term of the series, hence making the series as[tex]S=1+\sum_{r=1}^{\infty}\frac{(2r+1)^2}{(5^r)^2}[/tex]
    Now I have a hard time guessing what to do from now on. I expanded the numerator in summation series, 4r2 + 4r + 1. This formed the GP and AGP series (from 1 and 4r respectively). Now all that is left is to find the summation of 4r2/52r.

    By the way, I entered the series up at Wolfram|Alpha and the answer it showed is 5, which is correct.
     
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  3. May 26, 2013 #2

    mfb

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    There might be a direct way, but here is an indirect way:

    $$\sum_{r=1}^\infty \frac{r^2}{5^{2r}}\\
    = \frac{1}{25} + \sum_{r=2}^\infty \frac{r^2}{5^{2r}}\\
    = \frac{1}{25} + \sum_{r=1}^\infty \frac{(r+1)^2}{5^{2(r+1)}}\\
    = \frac{1}{25} + \frac{1}{25}\sum_{r=1}^\infty \frac{r^2+2r+1}{5^{2r}}\\
    = \frac{1}{25} \left(1+\sum_{r=1}^\infty \frac{2r+1}{5^{2r}}\right) + \frac{1}{25}\sum_{r=1}^\infty \frac{r^2}{5^{2r}}$$
     
  4. May 26, 2013 #3

    Ray Vickson

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    If you do know calculus, that summation would be a standard homework exercise. If
    [tex] S(x) = \sum_{n=1}^{\infty} x^n,[/tex]
    then
    [tex] x\frac{d}{dx} S(x) = \sum_{n=1}^{\infty} n x^n \\
    x \frac{d}{dx} \left[ x\frac{d}{dx} S(x) \right] = \sum_{n=1}^{\infty} n^2 x^n [/tex]
    You can find ##S(x)## in closed form, so you can do all the calculations explicitly. Then, of course, you set x = 1/5.
     
  5. May 27, 2013 #4

    AGNuke

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    Solution

    Thanks MFB. I think I got it.:approve: Here's the solution.[tex]S=1+4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}+4\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}[/tex]I can find the sum of an Arithmetic-Geometric Series, hence[tex]\sum_{r=1}^{\infty}\frac{r}{5^{2r}}=\frac{25}{576}[/tex]The Sum of third term, which is a simple Geometric Series,is[tex]\sum_{r=1}^{\infty}\frac{1}{5^{2r}}=\frac{1}{24}[/tex]The Problem was the first term, which I happened to resolve thanks to MFB.

    [tex]\frac{24}{25}\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{25}\left(1+2\sum_{r=1}^{\infty}\frac{r}{5^{2r}}+\sum_{r=1}^{\infty}\frac{1}{5^{2r}}\right)[/tex][tex]\Rightarrow 4\sum_{r=1}^{\infty}\frac{r^2}{5^{2r}}=\frac{1}{6}\left(1+2\times \frac{25}{576}+\frac{1}{24}\right)=\frac{325}{576}[/tex]Substituting All the known variables into the First Equation,[tex]S=\frac{1728}{1728}+\frac{325}{1728}+\frac{300}{1728}+\frac{72}{1728}= \frac{2425}{1728}[/tex][tex]\Rightarrow \frac{1728}{485}S=\frac{1728}{485}\times \frac{2425}{1728}=5[/tex]
    While I do know Calculus, I can only solve them when there is that Integral Sign all over, or in physics. Using it in this manipulative way is not something I've done before. But still, if you could show me what you actually meant, I can learn something, as I need to brace myself for my College Entrance Exam, The IIT-JEE which is only a week apart. So, a little more help will be appreciated.
     
    Last edited: May 27, 2013
  6. May 27, 2013 #5

    Ray Vickson

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    What I mean is just exactly what I said:
    [tex] \sum_{n=1}^{\infty} n x^n = x \frac{d}{dx} \sum_{n=1}^{\infty} x^n
    = x \frac{d}{dx} \frac{x}{1-x} = \frac{x}{(1-x)^2}[/tex]
    and
    [tex] \sum_{n=1}^{\infty} n^2 x^n = x \frac{d}{dx} \frac{x}{(1-x)^2}.[/tex]
    You can just go ahead and do the derivative.
     
  7. May 27, 2013 #6

    AGNuke

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    OK Thanks! Now I see... Thanks a bunch. I was forgetting the fact that S is the sum of an infinite converging GP.
     
    Last edited: May 27, 2013
  8. May 27, 2013 #7

    AGNuke

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    Now then this is alternate solution to find that term which was bothering me, courtesy Ray Vickson :approve:
    [tex]S=\sum_{r=1}^{\infty}x^{r}=\frac{x}{1-x}[/tex][tex]x\frac{\mathrm{d} S}{\mathrm{d} x}=\frac{x}{(1-x)^2}[/tex][tex]x\frac{\mathrm{d} }{\mathrm{d} x}\left (x\frac{\mathrm{d} S}{\mathrm{d} x} \right )=\frac{x(1+x)}{(1-x)^3}=\sum_{r=1}^{\infty}r^2x^r[/tex]

    Simply substituting ##x=\frac{1}{25}##, I got ##\frac{325}{1728\times 4}##, and multiplying it by 4 (as required) gives ##\frac{325}{1728}## which is the term I required.

    Really Interesting using differentiation to find the summation so easily. I definitely will try this method in further question I will attempt. :rolleyes:
     
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