# Ramanujan's sum - help with proof.

• August 6th 2010, 06:39 AM
Cairo
Ramanujan's sum - help with proof.
Hi

Let $c_{k}(n)=\underset{d|(n,k)}{\sum}d\mu(\frac{k}{d})
$

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
• August 6th 2010, 07:19 AM
chiph588@
$\displaystyle \sum_{d\mid k}d\cdot\mu\left(\frac kd\right)\left\lfloor\frac nd\right\rfloor = \sum_{d\mid k}\sum_{m\leq n/d} d\cdot\mu\left(\frac kd\right) = \sum_{\underset{d\leq k}{d\mid k}}\sum_{m\leq n/d} d\cdot\mu\left(\frac kd\right) = \sum_{m\leq n}\sum_{\underset{d\mid n}{d\mid k}}d\cdot\mu\left(\frac kd\right)$ where the last equality comes from a counting argument.

The last step is to realize $d\mid n, \; d\mid k \iff d\mid(n,k)$.
• August 6th 2010, 07:27 AM
Cairo
Thanks. I can see where I was going wrong.

I hate partial summations!!!