1. ## Two more questions

I will seperate my next 2 questions into 2 separate threads so that it does not get cluttered. I don't know if the following definition is needed, but I will include it.

$\displaystyle k$ is a positive integer and $\displaystyle f,g$ and $\displaystyle h$ are arithmetic functions. $\displaystyle \hat{f}$ denotes the Dirichlet series for $\displaystyle f$, so

$\displaystyle \hat{f}(s)$ = $\displaystyle \sum_{n=1}^{\infty}{f(n)/n^s}$

(1) Define $\displaystyle \lambda_k$=$\displaystyle \mu * u^{[k]}$. Show that

(a) If $\displaystyle k>1$ then for all primes $\displaystyle p$ and all $\displaystyle e$ $\displaystyle \in N:$

$\displaystyle \lambda_k{(p^e)}$=$\displaystyle \left\{\begin{array}{cc}1,&\mbox{ if }e \equiv 0 \mbox{ mod } k\\-1, & \mbox{ if }e \equiv 1 \mbox{ mod } k\\0, & \mbox{ otherwise } \end{array}\right.$

(b)$\displaystyle \widehat{\lambda_k}$(s)=$\displaystyle \frac{\zeta{(ks)}}{\zeta{(s)}}$

Any help would be appreciated, thanks!

2. It is: $\displaystyle u^{\left[ k \right]} \left( n \right) = \left\{ \begin{gathered} 1{\text{ if }}n = m^k {\text{ for some }}m \in \mathbb{Z}^ + \hfill \\ 0{\text{ otherwise}} \hfill \\ \end{gathered} \right.$ , right?

(a) $\displaystyle \lambda _k \left( {p^e } \right) = \sum\limits_{\left. d \right|p^e } {\mu \left( d \right) \cdot u^{\left[ k \right]} \left( {\tfrac{n} {d}} \right)} = \sum\limits_{j = 0}^e {\mu \left( {p^j } \right) \cdot u^{\left[ k \right]} \left( {p^{e - j} } \right)}$

Thus: $\displaystyle \lambda _k \left( {p^e } \right) = \mu \left( 1 \right) \cdot u^{\left[ k \right]} \left( {p^e } \right) + \mu \left( p \right) \cdot u^{\left[ k \right]} \left( {p^{e - 1} } \right) = u^{\left[ k \right]} \left( {p^e } \right) - u^{\left[ k \right]} \left( {p^{e - 1} } \right)$

Now consider the different cases
• If $\displaystyle e = \dot k$ we have $\displaystyle u^{\left[ k \right]} \left( {p^e } \right) = 1{\text{ and }}u^{\left[ k \right]} \left( {p^{e - 1} } \right) = 0 \Rightarrow \lambda _k \left( {p^e } \right) = 1$
• If $\displaystyle e\equiv{1}(\bmod.k)$ we have $\displaystyle u^{\left[ k \right]} \left( {p^e } \right) = 0{\text{ and }}u^{\left[ k \right]} \left( {p^{e - 1} } \right) = 1 \Rightarrow \lambda _k \left( {p^e } \right) = - 1$
• Otherwise neither $\displaystyle e$ nor $\displaystyle e-1$ is a multiple of k, thus $\displaystyle u^{\left[ k \right]} \left( {p^e } \right) = 0{\text{ and }}u^{\left[ k \right]} \left( {p^{e - 1} } \right) = 0 \Rightarrow \lambda _k \left( {p^e } \right) = 0$
(b)
Multiplying $\displaystyle \left( {\sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}} {{n^s }}} } \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{{\mu \left( n \right)}} {{n^s }}} } \right) = \sum\limits_{n = 1}^\infty {\tfrac{{\sum\limits_{\left. d \right|n} {\mu \left( d \right) \cdot u^{\left[ k \right]} \left( {\tfrac{n} {d}} \right)} }} {{n^s }}}$

Thus: $\displaystyle \left( {\sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}} {{n^s }}} } \right) \cdot \left( {\sum\limits_{n = 1}^\infty {\tfrac{{\mu \left( n \right)}} {{n^s }}} } \right) = \sum\limits_{n = 1}^\infty {\tfrac{{\lambda _k \left( n \right)}} {{n^s }}}$

Now by this post http://www.mathhelpforum.com/math-help/number-theory/33568-stumped-again.html
we can say that $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{u^{\left[ k \right]} \left( n \right)}} {{n^s }}} = \zeta \left( {s \cdot k} \right)$

Recall the identity: $\displaystyle \sum\limits_{n = 1}^\infty {\tfrac{{\mu \left( n \right)}} {{n^s }}} = \frac{1} {{\zeta \left( s \right)}}$

Therefore: $\displaystyle \sum\limits_{n = 1}^\infty \frac{\lambda_k \left( n \right)} {n^s } = \frac{{\zeta \left( {s \cdot k} \right)}} {{\zeta \left( s \right)}}$