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