Property of a divisor function, $\lim_{n\to\infty} \frac{\rho(n, n)}{n^2}=1-{\pi^2\over12}$ where $\rho(n, k):=\sum_{i=1}^k{(n\mod i)}$

Question

Let $$\rho(n, k):=\sum_{i=1}^k{(n\mod i)}$$ where $(n \mod i)$ is a remainder of $n$ divided by $k$. Also let $\sigma(n)$ be the sum of all factors of $n$, or $$\sigma(n):=\sum_{d|n} d$$ i.e. a division function of order 1. I found the relation between $\rho(n, n)$ and $\rho(n+1, n+1)$, which was $$\rho(n+1, n+1)=\rho(n, n)+2n+1-\sigma(n+1)$$ This follows because, for $i=1, 2, \cdots, n$, $((n+1)\mod i) \ \ \ \ -(n\ \ \mod i)$ is 1 if $i|n+1$ and $1-i$ otherwise, which in turn $$\rho(n+1, n+1)-\rho(n, n)=\sum_{ i\nmid n+1} 1+\sum_{i\mid n+1}{(1-i)}=\sum_{i\le n}1-\left(\sum_{i \mid n}i - (n+1)\right)\\=2n+1-\sigma(n+1)$$ Through this, I got the closed-form of $\rho(n, n)$, which was $$\rho(n, n)=n^2-1-\sum_{k=2}^n \sigma(k)=n^2-\sum_{k=1}^n \sigma(k)$$ Now, the real problem is, I have to show that $$\lim_{n\to\infty} \frac{\rho(n, n)}{n^2}=1-{\pi^2\over12}$$ and it gives a well-known property $\zeta(2)={\pi^2\over6}$ as a hint. It seems to prove $$\lim_{n\to\infty} \frac{\sum_{k=1}^n \sigma(k)}{n^2}={1\over2}\sum_{k=1}^\infty{1\over k^2}$$ but I have no idea. Any hints are going to be appreciated. Thanks.

Answer 1

For the sum of divisors, we have $$\sigma(m)=\sum_{kl=m}l,$$ where we sum over all pairs $(k,l)$ with $kl=m$. Thus, $$\sum^n_{m=1}\sigma(m)=\sum_{kl\le n}l,$$ the sum over all pairs of positive integers $(k,l)$ with $kl\le n$, i.e. under the hyperbola defined by $kl=n$. It's a two-dimensional sum, and if you sum over $k$, first, you get $$\sum_{kl\le m}l=\sum^n_{l=1}l\,\sum^{\lfloor n/l\rfloor}_{k=1}1=\sum^n_{l=1}l\,\lfloor n/l\rfloor,$$ interesting, but not helpful in this case. BTW, it's another proof of your result above, because $n\bmod l=n-l\lfloor n/l\rfloor$.

If you sum over $l$, first, you get $$\sum_{kl\le m}l=\sum^n_{k=1}\,\sum^{\lfloor n/k\rfloor}_{l=1}l=\sum^n_{k=1}\,\frac{\lfloor n/k\rfloor(\lfloor n/k\rfloor+1)}2.$$