Rational Ratios : a parametrized variant

You should specifiy what you mean by "the mean of the list $Q(k)$ ", which has no meaning in itself. Refering to the inspiration, let , for $x\in \mathbb{N}$ , $M_{x}={1,2,3,...,x}$ and $N_{x}(k)={k^1,k^2,..,k^x}$ . Then let $Q_{x}(k)=(\frac{i}{k^j})_{1\leq i\leq x,1\leq j\leq x}$ be the $x$ by $x$ matrix whose entries are the fractions you described whose numerators are less than $x$ and whose denominators are less than $k^x$ . Now the mean coefficient of this matrix, say $\mu_{x}(k)$ , is well-defined to be $\mu_{x}(k)=\frac{1}{x^2}\sum_{1\leq i,j \leq x}\frac{i}{k^j}$ $=\frac{x(x+1)}{2x^2}\sum_{1\leq j \leq x}\frac{1}{k^j}$ $=\frac{x(x+1)}{2x^2}(\frac{1-\frac{1}{k^{x+1}}}{k-1})\rightarrow_{x\rightarrow\infty} \frac{1}{2(k+1)}$ , which we define to be $S(k)$ .

Now, $T(n)=\sum_{k=2}^{\infty}n^{\frac{-1}{S(k)}}=\sum_{k\geq 1}\frac{1}{(n^2)^k}=\frac{1}{n^2-1}$ . From here, we compute $U=\sum_{r\geq 2} T(r)=\sum_{r \geq 2} \frac{1}{r^2-1}$ . First, let $U_{n}=\sum_{r=2}^{n}T(r)$ . Notice $\frac{1}{r^2-1}=\frac{1}{2}(\frac{1}{r-1}-\frac{1}{r+1})$ , so $U_{n}=\frac{1}{2}(\sum_{r=2}^{n}[\frac{1}{r-1}-\frac{1}{r+1}]=\frac{1}{2}(\sum_{r=1}^{n-1}\frac{1}{r}-\sum_{r=3}^{n+1}\frac{1}{r})=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1})\rightarrow_{n\rightarrow\infty}\frac{3}{4}$ . So $U=\frac{3}{4}$ , and $\sum_{p=1}^{\infty}p^{\frac{-8U}{3}}=\sum_{p=1}^{\infty}\frac{1}{p^2}=\zeta(2)$ , which everyone knows is $\frac{\pi^2}{6}$ .

$s(n,p)=\frac{(p (p+1)) \sum _{z=1}^p n^{-z}}{2 p^2}$ which is comparable to the $S(k)$ of the problem considering the as the dimension of the square matrix.

$\left( \begin{array}{cc} 1 & \infty \\ 2 & \frac{1}{2} \\ 3 & \frac{1}{4} \\ 4 & \frac{1}{6} \\ 5 & \frac{1}{8} \\ 6 & \frac{1}{10} \\ 7 & \frac{1}{12} \\ 8 & \frac{1}{14} \\ 9 & \frac{1}{16} \\ 10 & \frac{1}{18} \\ \end{array} \right)$

Therefore, a simpler limited form is $s(n)=\frac{1}{2 (n-1)}$ .

$t=\sum _{k=2}^{\infty } n^{-\frac{1}{s(k)}} \Rightarrow \frac{1}{n^2-1}$

$\sum _{n=2}^{\infty } t\Rightarrow \frac34$

$\sum _{p=1}^{\infty } p^{-\frac{8\ 3}{4\ 3}} \Rightarrow \frac{\pi ^2}{6} \approx 1.64493406684823$