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( Sum to infinity of a divisor function ) ^{2} simplify expression watch

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    Hi,

    I have  120 \sum \limits_1^\infty (\sigma_{3}(n))^{2} , where  \sigma_{3}(n) is the divisor function.

    And I want to show that this can be written as  120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k)

    I'm pretty stuck on ideas starting of to be honest, since the sum is infinite, any help much appreciated.

    Many thanks in advance.
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    (Original post by xfootiecrazeesarax)
    Hi,

    I have  120 \sum \limits_1^\infty (\sigma_{3}(n))^{2} , where  \sigma_{3}(n) is the divisor function.

    And I want to show that this can be written as  120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k)

    I'm pretty stuck on ideas starting of to be honest, since the sum is infinite, any help much appreciated.

    Many thanks in advance.
    Can't necessarily help, but I'm struggling to make sense of that.

    I presume \displaystyle\sigma_3(n)=\sum_{d  |n}d^3

    The sum you're given rapidly goes off to infinity, and you want to show it's equal to a a finite sum that is a function of n. :confused:
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    (Original post by ghostwalker)
    Can't necessarily help, but I'm struggling to make sense of that.

    I presume \displaystyle\sigma_3(n)=\sum_{d  |n}d^3

    The sum you're given rapidly goes off to infinity, and you want to show it's equal to a a finite sum that is a function of n. :confused:
    Agree with that, though, could it not be possible that n can go to infinity and this has just simplified an expression for each term?
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    (Original post by xfootiecrazeesarax)
    Agree with that, though, could it not be possible that n can go to infinity and this has just simplified an expression for each term?
    A quick check setting n=4 say, shows:

    \latex (\sigma_3(4))^2=(1+8+64)^2 = 73^2

    Whereas \sum\limits_{k=1}^3 \sigma_3(k)\sigma_3(4-k)=\sigma_3(1)\sigma_3(3)+\sigma  _3(2)\sigma_3(2)+\sigma_3(3)   \sigma_3(1)

    =1\times 28+ 9\times 9 +28\times 1=137\not=73^2
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    (Original post by xfootiecrazeesarax)

    And I want to show that this can be written as  120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k)
    From a bit of digging around here I came across:

    \sigma_7(n)-\sigma_3(n)= 120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k)

    though the level of maths is somewhat beyond me.

    Don't know if that's of any use, or might point you in a useful direction.
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    (Original post by ghostwalker)
    From a bit of digging around here I came across:

    \sigma_7(n)-\sigma_3(n)= 120\sum \limits_{k=1}^{n-1} \sigma_{3}(k) \sigma_{3}(n-k)
    .

    Yes, this is exactly what I am trying to show, since my question doesn't seem to make sense I will post the whole question instead to clarify things:

    I have concluded that E_{4}(t){2}=E_{8}(t) and I am wanting to use this to show that \sigma_{7}(n)=\sigma_{3}(n)+120 \sum\limits^{n-1}_{k=1} \sigma_{3}(k)\sigma_{3}(n-k)
    where  E_{8}(t)=1+480\sum\limits^{\inft  y}_{n=1} \sigma_{7}(n)q^{n}
    and E_{4}(t)=1+240\sum\limits^{\inft  y}_{n=1} \sigma_{3}(n)q^{n}
    And so my working so far is E_{4}^{2}(t)-E_{8}(t)=0
    (1+480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n} -((1+480\sum\limits^{\infty}_{n=1  } \sigma_{3}(n)q^{n})(1+480\sum\li  mits^{\infty}_{n=1} \sigma_{3}(n)q^{n}))  = 480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n} - 480\sum \limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}-240^{2}\sum\limits^{\infty}_{k=1  } \sigma_{3}(k)q^{k} \sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}
    Divide by 480 and everything looks on track except this issue in my OP \sigma_{7}(n)q^{n}=q^{n}\sigma_{  3}(n)+120\sum\limits^{\infty}_{k  =1} \sigma_{3}(k)q^{k} \sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}
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    (Original post by xfootiecrazeesarax)
    Yes, this is exactly what I am trying to show, since my question doesn't seem to make sense I will post the whole question instead to clarify things:

    I have concluded that E_{4}(t){2}=E_{8}(t) and I am wanting to use this to show that \sigma_{7}(n)=\sigma_{3}(n)+120 \sum\limits^{n-1}_{k=1} \sigma_{3}(k)\sigma_{3}(n-k)
    where  E_{8}(t)=1+480\sum\limits^{\inft  y}_{n=1} \sigma_{7}(n)q^{n}
    and E_{4}(t)=1+240\sum\limits^{\inft  y}_{n=1} \sigma_{3}(n)q^{n}
    And so my working so far is E_{4}^{2}(t)-E_{8}(t)=0
    (1+480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n} -((1+480\sum\limits^{\infty}_{n=1  } \sigma_{3}(n)q^{n})(1+480\sum\li  mits^{\infty}_{n=1} \sigma_{3}(n)q^{n}))  = 480\sum\limits^{\infty}_{n=1} \sigma_{7}(n)q^{n} - 480\sum \limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}-240^{2}\sum\limits^{\infty}_{k=1  } \sigma_{3}(k)q^{k} \sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}
    Divide by 480 and everything looks on track except this issue in my OP \sigma_{7}(n)q^{n}=q^{n}\sigma_{  3}(n)+120\sum\limits^{\infty}_{k  =1} \sigma_{3}(k)q^{k} \sum\limits^{\infty}_{n=1} \sigma_{3}(n)q^{n}
    As I said I'm not familiar with this, BUT it looks as if you have two series in q that are equal, and you just need to equate coefficients for q^n on either side of your E_4^2=E_8.
    EDit: So, there shouldn't be any infinite sums when you equate coeff.
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    (Original post by xfootiecrazeesarax)
    ...
    Yep, equating coeff. of q^n gives:

    480\sigma_3(n)+(240)^2\sum  \limits_{k=1}^{n-1}\sigma_3(k)\sigma_3(n-k) = 480\sigma_7(n)

    and divide by 480.
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    (Original post by ghostwalker)
    Yep, equating coeff. of q^n gives:

    480\sigma_3(n)+(240)^2\sum  \limits_{k=1}^{n-1}\sigma_3(k)\sigma_3(n-k) = 480\sigma_7(n)

    and divide by 480.

    okay cheers, I see that.

    However I'm struggling to write this neatly, so I've wrote out a bunch of terms and I can see the pattern, but my working is very improper, by any chance do you have any hints how to approach in a neat, proper manner?

    thanks .
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    (Original post by xfootiecrazeesarax)
    okay cheers, I see that.

    However I'm struggling to write this neatly, so I've wrote out a bunch of terms and I can see the pattern, but my working is very improper, by any chance do you have any hints how to approach in a neat, proper manner?

    thanks .
    I'm not really clear on what you're asking; can you explain a bit more?

    As regards, the working, I'd just say equating coefficients of q^n, we have, and the line I posted previously. I wouldn't add any intermediate steps - it's just consolidating the terms arising when two series are multiplied together.
 
 
 
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