Arithmetical function : How can I prove? How can I show that this sum $\sum_{d|n} \mu(d) \log^kd$ is $0$ where $\mu(d)$ is mobius function.
I've expect that this question is solved by induction..! 
$k$ is integer that is a power of $\log$
 A: Note that $$\sum_{d\mid n}\mu(d)\log^{k}(d)=\sum_{d\mid n}\mu\left(d\right)\left(-\log\left(\frac{n}{d}\right)+\log\left(n\right)\right)^{k}$$ $$=\sum_{m=0}^{k}\dbinom{k}{m}(-1)^{m}\log^{k-m}(n)\sum_{d\mid n}\mu(d)\log^{m}\left(\frac{n}{d}\right)
 $$ and $$\sum_{d\mid n}\mu(d)\log^{m}\left(\frac{n}{d}\right)=\Lambda_{m}\left(n\right)
 $$ is the generalized Von Mangoldt function. So we have that our sum is $$\sum_{d\mid n}\mu(d)\log^{k}(d)=\sum_{m=0}^{k}\dbinom{k}{m}(-1)^{m}\log^{k-m}(n)\Lambda_{m}\left(n\right).\tag{1}
 $$ Now it is well known that $\Lambda_{m}\left(n\right)=0
 $ iff $n$ has more than $m$ distinct prime factors, otherwise $\Lambda_{m}\left(n\right)>0
 $ (see for example here for a proof). So if we take $n
 $ such that it has more than $k$ distinct prime factors, the RHS of $(1)$ is trivially zero. If $n$ has less or equal than $k$ distinct prime factors the claim is false. You can find a counterexample rather easily.
A: For $k=1$ the identity should be the following
$$S_1(n)=\sum_{d|n}\mu(d)\ln(d)=\begin{cases}
0& \mbox{if $n$ is not a power of a prime},\\
-\ln(p)&\mbox{if $n$ is a power of a prime $p$.}
\end{cases}$$
Let $n=p_1^{a_1}\cdots p_r^{a_r}$ with $r>0$ then
$$S_1(n)=-\sum_{1\leq i_1\leq r}\ln(p_{i_1})+\sum_{1\leq i_1<i_2\leq r}(\ln(p_{i_1})+\ln(p_{i_2}))-\sum_{1\leq i_1<i_2<i_3\leq r}(\ln(p_{i_1})+\ln(p_{i_2})+\ln(p_{i_3}))+\cdots+ (-1)^r(\ln(p_{i_1})+\dots+\ln(p_{i_r})).$$
Now we count the terms $\ln(p_i)$ in the above sum
$$-1+(r-1)-\frac{1}{2}(r-1)(r-2)+\dots +(-1)^r=-\sum_{j=0}^{r-1} (-1)^{j}\binom{r-1}{j}\\=-(1-1)^{r-1}=\begin{cases}
0& \mbox{if $r>1$},\\
-1&\mbox{if $r=1$.}
\end{cases}$$
and we are done.
