Landau's theorem using nth roots This question was asked  earlier  at MSE .
Let $\omega$(n) denote the number of distinct primes dividing $n$. The Mobius function is defined as $\mu(n) = (-1)^{\omega(n)}$ if $n$ is squarefree and  $\mu(n) = 0$ otherwise .  Next let $S(n) = \sum_{k=1}^n \frac{\mu(k)}{k}$. It is known that $S(n)$ approaches zero as $n$ approaches infinity and that this is
equivalent to the prime number theorem (von Mangoldt, Landau). 
What happens if we replace powers of $(-1)$ in the Mobius function with other roots of unity?  To focus on a specific case, let's use fourth roots and 
define $f(n) = i^n$ if $n$ is squarefree and $f(n) = 0$ otherwise. Then $f(n)$ is a 
multiplicative function whose initial values are $(1,i,i,0,i,-1,i,0,0,-1,...)$. Finally, let $T(n) = \sum_{k=1}^n \frac{f(k)}{k}$. Does $T(n)$ have properties analogous to those of $S(n)$?  
Questions: $(1)$ Does $\sum_{k=1}^{\infty}$ $\frac{f(k)}{k}$ converge? [The corresponding infinite product $\prod_p (1 + i/p)$ does not converge since 
$\sum\frac{1}{p^2} < \infty$ while $\sum \frac{1}{p} = \infty$. ]
$(2)$ The partial sums $S(n)$ are known to satisfy $|S(n)| \leq 1$ for all $n$.
Are the partial sums $|T(n)|$ also bounded by some constant independent of $n$? [Over the initial stretch $1 \leq n \leq 20$ , one finds that the max
$T$ value is $|T(19)| = 1.57 ...$ ].
Thanks
 A: Theorem 1 in Section 6.1 of Tenenbaum's Introduction to Analytic and Probabilistic Number Theory states (after taking its $N=0$ for simplicity) that for any $|z|<2$,
$$
\sum_{n\le x} z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac z{p-1} \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}.
$$
The same method would show that
$$
\sum_{n\le x} \mu^2(n) z^{\omega(n)} \sim \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot x\, (\log x)^{z-1}.
$$
From here, partial summation gives (for $z\ne 0$)
$$
\sum_{n\le x} \mu^2(n) \frac{z^{\omega(n)}}n = \frac1{\Gamma(z)} \prod_p \bigg( 1 + \frac zp \bigg) \bigg( 1-\frac1p \bigg)^z \cdot \frac{(\log x)^z}z + c(z) + o(1)
$$
for some constant $c(z)$. So when $|z|<2$ and $\Re z<0$, the sum converges to this $c(z)$; when $|z|<2$ and $\Re z>0$, the sum grows to infinity in modulus. When $|z|<2$ and $\Re z=0$, the sum oscillates asymptotically around a circle in the complex plane. So the interesting case $z=i$, strangely, the answer to your question (1) is no while the answer to your question (2) is yes!
