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Let $\lambda(n)$ for integers $n\geq 1$ the Liouville function, see its definition for example from this Wikipedia. And we denote with $\Gamma(n)$ the particular values of the gamma function over positive integers $n\geq 1$, thus it is $(n-1)!$.

We consider $$\lim_{x\to-\infty}\sum_{n=1}^\infty\lambda(n)\frac{x^n}{\Gamma(n)}.\tag{1}$$ (Notice that is the limit of previous series as $x$ tends to $-\infty$).

Question. Is it possible to deduce that $(1)$ is finite? Many thanks.

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  • $\begingroup$ I hope that my question has mathematical meaning. I would like if it is possible to deduce that our limit is finite, is infinite or well oscillates. Many thanks all users. $\endgroup$ – user243301 Nov 21 '17 at 15:00
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    $\begingroup$ Let $f(x)=\sum_{n\geq 1}\lambda(n)\frac{(-1)^n x^n}{\Gamma(n)}$. We have $$ g(s)= (\mathcal{L} f)(s) = \sum_{n\geq 1}\frac{n\lambda(n) (-1)^n}{s^{n+1}} $$ and if we manage to prove that $\lim_{s\to 0^+} g(s)<0$ we have that $f(x)\to 0$ as $x\to +\infty$, since $f$ decays like $e^{-\lambda x}$ for some $\lambda>0$. If there is something known about the partial sums of $n\lambda(n)$ or $n\lambda(n)(-1)^n$, this is a good moment for exploiting it through summation by parts. It does not look to be a simple problem, anyway. $\endgroup$ – Jack D'Aurizio Nov 21 '17 at 20:16
  • $\begingroup$ Many thanks, one more time my problem has no more context (and my thougths were that maybe could have a suitable answer for the question), but it is good have your reasonings that I am going to read. I hope that other users want to do feedback about the question @JackD'Aurizio $\endgroup$ – user243301 Nov 21 '17 at 20:20
  • $\begingroup$ I promise a live goose as added prize for the first user solving this problem! (Not, it is a joke, in fact now I've no much money and each goose that I am meeting doesn't know how to fly yet.) $\endgroup$ – user243301 Nov 23 '17 at 20:12
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    $\begingroup$ I would also add that this answer of Peter Humphries on MO, together with my previous remark, points towards a positive answer to your conjecture, since it shows that $L(n)$ is predominantly negative. $\endgroup$ – Jack D'Aurizio Nov 23 '17 at 20:20
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Definitely too long for a comment.

Let $L(n)=\sum_{m)=1}^{n}\lambda(m)$. Quoting Peter Humphries from MO:

As for somewhat elementary methods (in the sense of avoiding the Riemann zeta function) to show that $L(n)$ is "usually" of order $\sqrt{n}$, one can use the Lambert series $$\sum_{n=1}^{\infty}{\frac{\lambda(n)q^n}{1-q^n}} = \sum_{n=1}^{\infty}{q^{n^2}}.$$ As $$\frac{q^n}{1+q^n} = \frac{q^n}{1-q^n} - 2\frac{q^{2n}}{1-q^{2n}},$$ we have $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{q^{-n}+1}} = \sum_{n=1}^{\infty}{q^{n^2}} - 2\sum_{n=1}^{\infty}{q^{2n^2}}$$ or equivalently, letting $q = e^{-\pi/x}$ and $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}}$, where $x$ is large, $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \psi(1/x) - 2\psi(2/x)$$ Now $\psi(x)$ satisfies the functional equation $$\frac{1+2\psi(x)}{1+2\psi(1/x)} = \frac{1}{\sqrt{x}},$$ and so we can rewrite this as $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x}.$$ For large $x$, the left-hand side "looks like" $L(x)$, whereas the right-hand side is dominated by the term $\frac{1-\sqrt{2}}{2}\sqrt{x}$. This also explains why $L(n)$ is predominantly negative, as $\frac{1-\sqrt{2}}{2}$ is negative.

Now comes my comment under the main thread (with a $\color{blue}{\text{blue}}$ addendum):

Let $f(x)=\sum_{n\geq 1}\lambda(n)\frac{(-1)^n x^n}{\Gamma(n)}$. We have $$ g(s)= (\mathcal{L} f)(s) = \sum_{n\geq 1}\frac{n\lambda(n) (-1)^n}{s^{n+1}} =\color{blue}{-\frac{d}{ds}\sum_{n\geq 1}\frac{\lambda(n)(-1)^n}{s^n}}$$ and if we manage to prove that $\lim_{s\to 0^+} g(s)<0$ we have that $f(x)\to 0$ as $x\to +\infty$, since $f$ decays like $e^{-\lambda x}$ for some $\lambda>0$. If there is something known about the partial sums of $n\lambda(n)$ or $n\lambda(n)(-1)^n$, this is a good moment for exploiting it through summation by parts. It does not look to be a simple problem, anyway.

Not a simple problem, indeed, but to remark the connection between $\mathcal{L} f$ and the Jacobi $\Theta$ function/the Jacobi triple product is enlightening. We have $$ \sum_{n\geq 1}\frac{\lambda(n)(-1)^n}{s^n} = -\sum_{\substack{n\geq 1\\ n\text{ odd}}}\left(\frac{\lambda(n)}{s^n}+\frac{\lambda(n)}{s^{2n}}-\frac{\lambda(n)}{s^{4n}}+\frac{\lambda(n)}{s^{8n}}-\ldots\right)$$ and by setting $s=e^{\pi/x}$ (for $x\to +\infty$) the behaviour of the RHS (and its derivative) in a right neighbourhood of $s=0$ can be studied through Humphries' functional equation. I still have to complete the the necessary estimations in full rigor, but this points towards an affirmative answer to your question.

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  • $\begingroup$ You take great dedication in your answers, many thanks today I am going to study your answer and Humphries' result. $\endgroup$ – user243301 Nov 24 '17 at 4:58

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