The functions in question are

$$L(s)=\sum_{k=1}^\infty \frac{\lambda(k)}{k^s}=\frac{\zeta(2s)}{\zeta(s)} \mbox{ and } L^*(s)=\frac{1}{2}\sum_{k=1}^\infty \frac{\lambda(k)+(-1)^{k+1}}{k^s}=\frac{L(s)+\eta(s)}{2},$$

where

- $\lambda(k) = (-1)^{\Omega(k)}$ is the Liouville function
- $\Omega(k)$ is the prime omega function, counting the number of prime factors of $k$ including multiplicity
- $\zeta$ is the Riemann zeta function and $\eta$ is the Dirichlet eta function
- $s=\sigma + it$ is a complex number
- $p_1, p_2,\dots$ denote the primes with $p_1=2$

An equivalent discussion could be based on the sister Moebius and Merterns function, but here our focus is on Liouville. The notations $L_n$ and $L_n^*$ are used to denote the first $n$ terms of the series representing respectively $L(s)$ and $L^*(s)$ when $s=0$. The asymptotic behavior of $L_n$ has well-known deep implications on the Riemann Hypothesis (RH), discussed later in this post.

**Note**: the series defining $L^*(s)$ might or might not converge if $\frac{1}{2}<s\leq 1$ (that's the purpose of my question), however the formula $L^*(s)=(L(s)+\eta(s))/2$ is valid only for $\Re(s)>1$ because the series defining $L(s)$ converges only for $\Re(s)>1$ (at least according to Wikipedia, see here).

My main question can be stated in simple words:

**My question**

Does the series representing $L^*(s)$ converge and is it an analytic function if $\Re(s)>\frac{1}{2}$? Probabilistic arguments in favor of a positive answer are discussed in the appendix. Assuming the answer is yes, then $L(s)=2L^*(s)-\eta(s)$ is an analytic continuation of the original $L(s)$, from $\Re(s)>1$ to $\Re(s)>\frac{1}{2}$. Consequences are discussed in the next section.

I computed $L^*(s)$ for various $s$ with $\frac{1}{2}<\Re(s)\leq 1$ using the first million terms of the series, and compared with $L^*(s)=\frac{1}{2}\Big(\eta(s)+\frac{\zeta(2s)}{\zeta(s)}\Big)$ computed by Mathematica: the first three digits are identical. In particular, $L^*(1)=\frac{\log 2}{2}$ and the analytic continuation gives $L(1)=0$.

**Discussion**

The sequence $\{\lambda(k)\}$ consists of $+1$ and $-1$ that seem rather randomly distributed, though not perfectly randomly, but supposedly randomly enough (see here) as to imply the Riemann Hypothesis (RH) - still a conjecture at this point. For instance, if you look at runs of $+1$ or $-1$ (subsequences of consecutive $+1$ or consecutive $-1$) the probability for a run to be of length $m>0$ is equal to $2^{-m}$, as in a sequence of i.i.i. Bernoulli trials. More about this is discussed in the appendix. Yet this is not enough to make the series for $L(s)$ converge if $\Re(s)\leq 1$. Also, it seems there is a little bias in the sequence $\{\lambda(k)\}$, which might favor $L_n$ to be negative more frequently than positive, unlike perfect random walks. For instance, if $1<n< 906150257$, then $L_n$ is always negative (see here), but not at $n=906380357$, thus disproving Polya's conjecture. That bias might disappear when $n$ becomes extremely large.

Now if you replace $\lambda(k)$ by $\lambda^*(k)=(\lambda(k)+(-1)^{k+1})/2$ corresponding to the terms in the series defining $L^*(s)$, the distribution of run lengths (if you ignore the terms $\lambda^*(k)$ equal to zero), is significantly altered. This would also be true, with the exact same impact, if you apply the same trick to i.i.d. Bernoulli trials or other sequences with similar distribution of $+1$ and $-1$. In the sequence $\{\lambda^*(k)\}$, assuming you omit the zero terms which do not contribute to the sums $L_n^*$ or $L^*(s)$, the probability for a run to be of length $m>0$ is now equal to $2\cdot 3^{-m}$, down from $2^{-m}$ for the original sequence $\{\lambda(k)\}$. This means that on average, runs are now much shorter, to the point that it makes the series $L^*(s)$ converge even if $\frac{1}{2}<\Re(s)\leq 1$. In essence, our trick is what gets us an analytic continuation from $\Re(s)>1$ to $\Re(s)>\frac{1}{2}$. Transforming $L(s)$ into $L^*(s)$ is the same trick as transforming $\zeta(s)$ into $\eta(s)$, though in the latter case, it extents analycity from $\Re(s)>1$ to $\Re(s)>0$.

An interesting question is whether you can infer some useful result about $\lim \sup_{n \rightarrow \infty} L_n$ from the smoother $L^*_n$, as the $\lim\sup$ in question is intimately connected to the Riemann Hypothesis (see my previous post here). Another curious facts is:

$$\sum_{k=1}^n \lambda(k)\Big\lfloor \frac{n}{k}\Big\rfloor =\lfloor \sqrt{n}\rfloor ,$$

where $\lfloor \cdot \rfloor$ denotes the integer part function. This formula is mentioned here, and it allows you to compute $\lambda(k)$ recursively without using a table of prime numbers. Finally,

$$\zeta(s)=\frac{\zeta(2s)}{2L^*(s)-\eta(s)} = \frac{1}{2L^*(s)-\eta(s)}\cdot\prod_{k=1}^\infty \frac{1}{L(2^k s)}$$

gives (by successive applications of the first equality) an infinite product for $\zeta(s)$ converging (and analytic) if $\Re(s)>\frac{1}{2}$. The roots of $\zeta(s)$ in the critical strip $\frac{1}{2}<\Re(s)<1$ (if any, RH says that there are none) are identical to the poles of $L^*(s)$, or in other words, to the roots of $1/L^*(s)$ in the same strip.

**Appendix**

Here we show what would happen if the numbers $\lambda(k)$ were replaced by independent random variables $X_k$ taking value $1$ with probability $\frac{1}{2}$, and $-1$ with the same probability. This is pretty much (but not exactly) the way the $\lambda(k)$'s are behaving. Let's define

$$Z=\sum_{k=1}^\infty \frac{X_k}{k^s}.$$

Here $Z$ is a complex random variable, and $s=\sigma +it$. Its expectation is zero, and its variance is given by

$$Var[\Re(Z)]=Var[X_1]\cdot\sum_{k=1}^\infty \frac{\cos^2(t\log k)}{k^{2\sigma}},\\ Var[\Im(Z)]=Var[X_1]\cdot\sum_{k=1}^\infty \frac{\sin^2(t\log k)}{k^{2\sigma}}.$$

Both series converge if $\sigma=\Re(s)>\frac{1}{2}$. In that case $Var[Z]=Var[\Re(Z)]+Var[\Im(Z)] =Var[X_1]\cdot \zeta(2\sigma)$. A formula related to the characteristic function, if $\tau$ is a real number, is the following:

$$E[\exp(i\tau Z)]=\prod_{k=1}^\infty \cos(\tau k^{-s}) .$$

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