analytic-number-theoryfibonacci-numbersnt.number-theory
Helo,
Let $F(n)$ be the $n$th Fibonacci number, if $\left\{ x\right\}$ denotes the fractional part of $x$, how proving
$$\lim_{n\rightarrow\infty}\frac{1}{2n}\sum_{k=1}^{2n}\left\{ \frac{F(2n)}{F(k)}\right\} =1-\log\left(\frac{1+\sqrt{5}}{2}\right)$$
A lecturer wrote this formula on the board without reference and without proof! Thank you for any hint.
ps: the proposed limit is wrong. See the answer below which proves that the limit does indeed exist but has value $3\log(2)/4$.
This is not a complete answer but hopefully provides some insight.
The function $$\exp\left(-\sum\limits_{p\le x},\frac{\cos(x\, \log(p))}{\sqrt{p}}\right)\tag{1}$$
seems to approximate
$$\left|\frac{1}{\zeta\left(\frac{1}{2}+i x\right)}\right|\tag{2}$$
which is illustrated in Figure (1) below.
Figure (1): Illustration of formulas (1) and (2) in orange and blue
The sum
$$-\sum\limits_{\rho} \frac{\cos\left(\Im(\rho)\, \log(x)\right)}{\Im(\rho)}\tag{3}$$
seems to be more related to the derivative of the explicit formula for
$$\psi(x)=\sum\limits_{n=1}^x \lambda(n)\tag{4}$$
which is related to the Dirichlet series
$$-\frac{\zeta'(s)}{\zeta(s)}=\sum\limits_{n=1}^x \frac{\lambda(n)}{n^s}\,,\quad\Re(s)>1\tag{5}.$$
More specifically the explicit formula for $\psi(x)$ contains the following sum over the non-trivial zeta zeros
$$-\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \left(\frac{x^{\rho_{-k}}}{\rho_{-k}}+\frac{x^{\rho_k}}{\rho_k}\right)\right)\tag{6}$$
with derivative
$$-\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \left(x^{\rho_{-k}-1}+x^{\rho_k-1}\right)\right)\tag{7}$$
which assuming the Riemann hypothesis can be evaluated as
$$-\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \frac{2 \cos\left(\Im\left(\rho_k\right)\, \log(x)\right)}{\sqrt{x}}\right)\tag{8}.$$
If you look at your second plot carefully you'll see there seems to be peaks at the prime-powers as well as the primes which is consistent with the integers where $\psi(x)$ takes a step.
Best Answer
The statement is not correct. The correct statement is: $$\lim_{n\rightarrow\infty}\frac{1}{2n}\sum_{k=1}^{2n}\left\{ \frac{F(2n)}{F(k)}\right\}=\frac{3\log 2}{4}.$$ That is, the limit is $\approx 0.519860385$ instead of $\approx 0.518788175$. As a numerical verification, for $n=10^6$, the average under the limit is $\approx 0.519853615$ (according to SAGE).
More generally, we shall prove the following. $$\sum_{k=1}^n\left\{\frac{F(n)}{F(k)}\right\}= \begin{cases} \frac{\pi}{8}n+O(\sqrt{n}),&\text{$n$ is odd};\\ \frac{3\log 2}{4}n+O(\sqrt{n}),&\text{$n$ is even}. \end{cases}$$ The idea of the proof is that the fractional part of $F(n)/F(k)$ is typically very close to $0$ or $1$, and the two cases follow a simple pattern. We can restrict to $k\nmid n$ in the sum, because otherwise the fractional part is zero. Accordingly, we shall consider $$n=2mk\pm r\qquad\text{with}\qquad r\in\{1,2,\dotsc,k-1\}.$$
If $L(v)$ denotes the $v$-th Lucas number, then we have the identity $$F(u)L(v)=\begin{cases} F(u+v)+(-1)^v F(u-v),&u\geq v;\\ F(u+v)-(-1)^u F(v-u),&u\leq v. \end{cases}$$ Plugging $(mk,mk\pm r)$ for $(u,v)$, we obtain the following congruences: \begin{alignat*}{2} F(2mk+r)&\equiv F(r)(-1)^{mk}&&\pmod{F(mk)},\\ F(2mk-r)&\equiv F(r)(-1)^{mk-r-1}&&\pmod{F(mk)}. \end{alignat*} These congruences are also valid modulo $F(k)$, because $F(k)$ divides $F(mk)$. We infer that: $$\{F(n)/F(k)\}= \begin{cases} F(r)/F(k),&n=2mk+r,\quad mk\equiv 0\pmod{2};\\ 1-F(r)/F(k),&n=2mk+r,\quad mk\not\equiv 0\pmod{2};\\ 1-F(r)/F(k),&n=2mk-r,\quad mk\equiv n\pmod{2};\\ F(r)/F(k),&n=2mk-r,\quad mk\not\equiv n\pmod{2}.\\ \end{cases}$$ It is now convenient to introduce the notation $$I(n,t):=\mathbb{N}\cap\left(\frac{n}{t+1},\frac{n}{t}\right),$$ because then the cases $n=2mk+r$ correspond to $k\in I(n,2m)$, while the cases $n=2mk-r$ correspond to $k\in I(n,2m-1)$. Then we see that:
Now we choose a positive integer $M$, and sum up the contributions of $I(n,2m)$ and $I(n,2m-1)$ for $m\in\{1,2,\dotsc,M\}$. This way we see that $$\sum_{k=\lceil n/(2M+1)\rceil}^{n}\left\{\frac{F(n)}{F(k)}\right\}=\sum_{m=1}^M\sum_{k\in I(n,2m)}\left\{\frac{F(n)}{F(k)}\right\} +\sum_{m=1}^M\sum_{k\in I(n,2m-1)}\left\{\frac{F(n)}{F(k)}\right\}.\tag{$\ast$}$$ If $n$ is odd, then up to an error of $O(M)$, the right-hand side of $(\ast)$ equals $$\frac{1}{2}\sum_{\substack{1\leq m\leq M\\\text{$m$ odd}}}\left(\frac{n}{2m-1}-\frac{n}{2m+1}\right)=\frac{\pi}{8}n+O\left(\frac{n}{M}\right).$$ The contribution of $k<n/(2M+1)$ is $O(n/M)$, hence in fact $$\sum_{k=1}^n\left\{\frac{F(n)}{F(k)}\right\}=\frac{\pi}{8}n+O\left(M+\frac{n}{M}\right).$$ If $n$ is even, then up to an error of $O(M)$, the right-hand side of $(\ast)$ equals $$\frac{1}{2}\sum_{\substack{1\leq m\leq M\\\text{$m$ odd}}}\left(\frac{n}{2m-1}-\frac{n}{2m+1}\right)+\sum_{\substack{1\leq m\leq M\\\text{$m$ even}}}\left(\frac{n}{2m-1}-\frac{n}{2m}\right)=\frac{3\log 2}{4}n+O\left(\frac{n}{M}\right).$$ The contribution of $k<n/(2M+1)$ is $O(n/M)$, hence in fact $$\sum_{k=1}^n\left\{\frac{F(n)}{F(k)}\right\}=\frac{3\log 2}{4}n+O\left(M+\frac{n}{M}\right).$$ In both cases we choose $M=\lfloor\sqrt{n}\rfloor$, and we obtain the claimed asymptotic formula with error term $O(\sqrt{n})$.