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