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It's a well known result that the Dirichlet series of the Liouville function $ \lambda(n) $ is given by

$$ \sum_{k=1}^{\infty} \frac{\lambda(k)}{k^s} = \frac{\zeta(2s)}{\zeta(s)} $$

If we use Perron's formula on this, we get some expression like

$$ L(x) = \sum_{k \leq x} \lambda(k) = \frac{x^{1/2}}{\zeta(1/2)} + \sum_{\rho} \frac{x^\rho}{\rho} \frac{\zeta(2 \rho)}{\zeta'(\rho)} + O(1) $$

where the sum runs over the nontrivial zeroes (of multiplicity one) of the Riemann zeta function. Since $ \zeta(1/2) = -1.46 \ldots $, if we assume the Riemann hypothesis this suggests that the function $ L $ should have a bias to be negative, and this is indeed true: it's the subject of the Polya conjecture, for instance. Moreover, the first term of this expression also predicts the magnitude of the bias quite well: just to give an example, $ L(10^4) = -94 $, and based just on the first term of the above expression we would predict it's around $ -100/1.46 \approx -68 $, which is not far off from the actual value.

My question is this: often we can get "elementary intuition" into the behavior of these kinds of sums by pseudorandomness heuristics, appropriately modified. For example, the Riemann hypothesis is (roughly) equivalent to the central limit theorem bound on the function $ L(x) $. Is there such an intuition for why the partial sums of the Liouville function would be predominantly negative? Any intuition would probably have to contrast the Liouville function with the Mobius function, since the absence of the $ \zeta(2s) $ on the numerator from the Dirichlet series of $ \mu(k) $ means that the partial sums of the Mobius function show no such bias to be negative.

I would be happy with any answer which doesn't appeal to some analytic number theoretic machinery: Perron's formula, Mellin transforms, et cetera. Ideally I'm looking for something of the form "there are slightly more primes that are $ 3 \pmod{4} $ than primes $ 1 \pmod{4} $ because the congruence class $ 1 \pmod{4} $ has all of the odd squares in it".

Edit: There has been another question on this subject a decade ago, but it got no answer of the nature I'm looking for here.

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    $\begingroup$ Does this answer your question? Heuristic reason for Polya's conjecture $\endgroup$
    – Peter Humphries
    Commented Feb 16, 2022 at 16:58
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    $\begingroup$ @PeterHumphries I saw your question before asking mine, but your question didn't get any answer of the kind I wanted: the only answer is just saying "if we use Perron's formula and assume the zeroes are behaving nicely, then the bias coming from the pole at $ s = 1/2 $ survives". I agree it's kind of a duplicate otherwise. $\endgroup$
    – Ege Erdil
    Commented Feb 16, 2022 at 17:00
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    $\begingroup$ We have $\sum_{k \leq x } \lambda(k) = \sum_{k\leq x} \mu(k) \left\lfloor \sqrt{ \frac{x}{k}} \right\rfloor $, and removing the floor function (i.e. assuming the fractional part is uncorrelated with $\mu(k)$ ) gives $ \sqrt{x} \sum_{k\leq x} \frac{\mu(k)}{ \sqrt{k}} $ which is a partial sum of the zeta function. So there might not be a solution without discussing the zeta function. $\endgroup$
    – Will Sawin
    Commented Feb 16, 2022 at 17:27
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    $\begingroup$ $\sum_{k\leq x} \frac{\mu(k)}{\sqrt{k}}$ is a partial sum of the series $\sum_{k} \frac{\mu(k)}{\sqrt{k}}$, which formally evaluates to $\zeta(1/2)^{-1}$, and I think the partial sums converge to $\zeta(1/2)^{-1}$ if RH is true. $\endgroup$
    – Will Sawin
    Commented Feb 16, 2022 at 20:41
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    $\begingroup$ @WillSawin $\sum \mu(n)/n^s$ does not converge at $s = 1/2$, or in fact anywhere on ${\rm Re}(s) = 1/2$: if it converged at a point on that line then $\sum \mu(n)/n^s = 1/\zeta(s)$ for ${\rm Re}(s) > 1/2$ and we know $1/\zeta(s)$ has poles on the line ${\rm Re}(s) = 1/2$. When a Dirichlet series $\sum a_n/n^s$ converges on a half-plane ${\rm Re}(s) > \sigma_0$ and has a meromorphic continuation to some line segment on ${\rm Re}(s) = \sigma_0$ that includes a pole, it doesn't converge anywhere on that line. This also explains why $\sum 1/n^s$ doesn't converge anywhere on ${\rm Re}(s)=1$. $\endgroup$
    – KConrad
    Commented Jun 26, 2022 at 16:43

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