Cantor's Serpent

Since the question keeps changing, let us prove once and for all that $\lim_{n\to\infty}\frac{\mu(a_i,\frac{n(n+1)}2)}{n\ln n}=\frac12.$

Let $S_n$ be the sum of the first $\frac{n(n+1)}2$ terms; we wish to find $\lim_{n\to\infty}\frac{S_n}{n^2\ln n}$ . Firstly, $\begin{aligned} S_n&=\frac11+\frac12+\cdots+\frac1{n-1}+\frac1n\\ &\quad+\frac21+\frac22+\cdots+\frac2{n-1}\\ &\quad+\cdots\\ &\quad+\frac n1\\ &=H_n+2H_{n-1}+\cdots+(n-1)H_2+nH_1, \end{aligned}$ where $H_k=1+\frac12+\cdots+\frac1k$ .

Define $\epsilon_k=H_k-\ln k$ . Then $S_n=T_n+\epsilon_n+2\epsilon_{n-1}+\cdots+n\epsilon_1,$ where $T_n=\ln n+2\ln(n-1)+\cdots+(n-1)\ln2+n\ln1$ .

Recall the standard result $\lim_{k\to\infty}\epsilon_k=\gamma=0.577\ldots$ . Hence the sequence $|\epsilon_k|$ is bounded by an absolute constant $E$ , so $\left|\frac{\epsilon_n+2\epsilon_{n-1}+\cdots+n\epsilon_1}{n^2\ln n}\right|\leq\frac{\frac{n(n+1)}2E}{n^2\ln n}\to0$ as $n\to\infty$ . Hence we are left to find $\lim_{n\to\infty}\frac{T_n}{n^2\ln n}$ .

Now note that $\begin{aligned} T_n&=(\ln\frac nn+\ln n)+2(\ln\frac{n-1}n+\ln n)+\cdots+n(\ln\frac1n+\ln n)\\ &=\frac{n(n+1)}2\ln n+\sum_{j=1}^n f(\frac jn)+n\sum_{j=1}^n g(\frac jn), \end{aligned}$ where $f(x)=\ln x$ and $g(x)=(1-x)\ln x$ .

But as $n\to\infty$ , we have $\frac1n\sum_{j=1}^n f(\frac jn)\to\int_0^1\!\ln x\,dx=-1,$ and $\frac1n\sum_{j=1}^n g(\frac jn)\to\int_0^1\!(1-x)\ln x\,dx=-\frac34.$

Hence $\begin{aligned} \lim_{n\to\infty}\frac{T_n}{n^2\ln n} &=\lim_{n\to\infty}\left(\frac{\frac{n(n+1)}2\ln n}{n^2\ln n}+\frac{\frac1n\sum_{j=1}^n f(\frac jn)}{n\ln n}+\frac{\frac1n\sum_{j=1}^n g(\frac jn)}{\ln n}\right)\\ &=\lim_{n\to\infty}\left(\frac{\frac{n(n+1)}2\ln n}{n^2\ln n}-\frac1{n\ln n}-\frac{\frac34}{\ln n}\right)\\ &=\frac12-0-0\\ &=\frac12, \end{aligned}$ as desired.