# Thread: Ramanujan's sum - help with proof.

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

Hi

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

I am trying to prove that

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

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

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

Thanks

2. $\displaystyle \displaystyle \sum_{d\mid k}d\cdot\mu\left(\frac kd\right)\left\lfloor\frac nd\right\rfloor = \sum_{d\mid k}\sum_{m\leq n/d} d\cdot\mu\left(\frac kd\right) = \sum_{\underset{d\leq k}{d\mid k}}\sum_{m\leq n/d} d\cdot\mu\left(\frac kd\right) = \sum_{m\leq n}\sum_{\underset{d\mid n}{d\mid k}}d\cdot\mu\left(\frac kd\right)$ where the last equality comes from a counting argument.

The last step is to realize $\displaystyle d\mid n, \; d\mid k \iff d\mid(n,k)$.

3. Thanks. I can see where I was going wrong.

I hate partial summations!!!