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Math Help - Ramanujan's sum - help with proof.

  1. #1
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    Ramanujan's sum - help with proof.

    Hi

    Let c_{k}(n)=\underset{d|(n,k)}{\sum}d\mu(\frac{k}{d})<br />

    I am trying to prove that

    \underset{m=1}{\overset{n}{\Sigma}}c_{k}(m)=\under  set{d|k}{\Sigma}d\mu(\frac{k}{d})[\frac{n}{d}].

    This is what I have done, but I am not sure it is completely correct. Is anybody able to offer some advice?

    \underset{m=1}{\overset{n}{\Sigma}}c_{k}(m)=\overs  et{n}{\underset{m=1}{\Sigma}}\underset{d|(n,k)}{\S  igma}d\mu(\frac{k}{d})=\underset{d|n}{\Sigma}\unde  rset{\underset{d|k}{k=1}}{\overset{n}{\Sigma}}\mu(  \frac{k}{d})=\underset{d|n}{\Sigma}d\underset{k\le  q n}{\Sigma}\mu(\frac{k}{d})=\underset{d|k}{\Sigma}d  \mu(\frac{k}{d})[\frac{n}{d}]

    Thanks
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  2. #2
    MHF Contributor chiph588@'s Avatar
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     \displaystyle \sum_{d\mid k}d\cdot\mu\left(\frac kd\right)\left\lfloor\frac nd\right\rfloor = \sum_{d\mid k}\sum_{m\leq n/d} d\cdot\mu\left(\frac kd\right) = \sum_{\underset{d\leq k}{d\mid k}}\sum_{m\leq n/d} d\cdot\mu\left(\frac kd\right) = \sum_{m\leq n}\sum_{\underset{d\mid n}{d\mid k}}d\cdot\mu\left(\frac kd\right) where the last equality comes from a counting argument.

    The last step is to realize  d\mid n, \; d\mid k \iff d\mid(n,k) .
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  3. #3
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    Thanks. I can see where I was going wrong.

    I hate partial summations!!!
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