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Sums of this sort are well known to the experts - but since none of them have answered so far, let me try. Denoting the fractional part of a real $x$ by ${x}$, \{x\}$, you can write your sum as $$S = n^2 \sum_{d=1}^n \frac1{d^2} - 2n \sum_{d=1}^n \frac1d\left\{\frac nd\right\} + \sum_{d=1}^n \left\{\frac nd\right\}^2 = \frac{\pi^2}{6}\, n^2 + O(n\log n).$$ If you need more precision, you have to find the main term of the sum$\sum_d d^{-1}\{d^{-1}n\}$. The standard technique here, I believe, would be to use the Fourier expansion of the fractional part function, but you'd better contact experts for details, to avoid re-inventing the wheel. Here is a different kind of answer, depending on what you are after. Your sum counts the number of triples$d,x,y\in[1,n]$with$xd,yd\le n$. Since there are$\sum_{k=1}^n \tau(k)$such triples with$x=y$, splitting the sum into two part according to whether$x\ge y$or$y\ge x$, we can write it as $$S = 2 \sum_{dx\le n} x - \sum_{k=1}^n \tau(k).$$ Letting$k=dx$, we get $$S = 2 \sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k),$$ where$\sigma$is the sum-of-divisors function. This gives you an identity containing arithmetic functions'', as you requested. 3 edited body; added 4 characters in body; deleted 653 characters in body Sums of this sort are well known to the experts - but since none of them have answered so far, let me try. Denoting the fractional part of a real$x$by${x}$, you can write your sum as $$S = n^2 \sum_{d=1}^n \frac1{d^2} - 2n \sum_{d=1}^n \frac1d\left\{\frac nd\right\} + \sum_{d=1}^n \left\{\frac nd\right\}^2 = \frac6{\pi^2} n^2 + O(n\log n).$$ If you need more precision, you have to find the main term of the sum$\sum_d d^{-1}\{d^{-1}n\}$. The standard technique here, I believe, would be to use the Fourier expansion of the fractional part function, but you'd better contact experts for details, to avoid re-inventing the wheel. Sums of this sort are well known to the experts - but since none of them have answered so far, let me try. Denoting the fractional part of a real$x$by${x}$, you can write your sum as $$S = n^2 \sum_{d=1}^n \frac1{d^2} - 2n \sum_{d=1}^n \frac1d\left\{\frac nd\right\} + \sum_{d=1}^n \left\{\frac nd\right\}^2 = \frac6{\pi^2} frac{\pi^2}{6}\, n^2 + O(n\log n).$$ If you need more precision, you have to find the main term of the sum$\sum_d d^{-1}\{d^{-1}n\}$. The standard technique here, I believe, would be to use the Fourier expansion of the fractional part function, but you'd better contact experts for details, to avoid re-inventing the wheel. Here is a different kind of answer, depending on what you are after. Your sum counts the number of triples$d,x,y\in[1,n]$with$xd,yd\le n$. Since there are$\sum_{k=1}^n \tau(k)$such triples with$x=y$, splitting the sum into two part according to whether$x\ge y$or$y\ge x$, we can write it as $$S = 2 \sum_{dx\le n} x - \sum_{k=1}^n \tau(k).$$ Letting$k=dx$, we get $$S = 2 \sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k),$$ where$\sigma$is the sum-of-divisors function. This gives you an identity containing arithmetic functions'', as you requested. 2 added 1247 characters in body Sums of this sort are well known to the experts - but since none of them have answered so far, let me try. Denoting the fractional part of a real$x$by${x}$, you can write your sum as $$S = n^2 \sum_{d=1}^n \frac1{d^2} - 2n \sum_{d=1}^n \frac1d\left\{\frac nd\right\} + \sum_{d=1}^n \left\{\frac nd\right\}^2 = \frac6{\pi^2} n^2 + O(n\log n).$$ If you need more precision, you have to find the main term of the sum$\sum_d d^{-1}\{d^{-1}n\}$. The standard technique here, I believe, would be to use the Fourier expansion of the fractional part function, but you'd better contact experts for details, to avoid re-inventing the wheel. Here is a different kind of answer, depending on what you are after. Your sum counts the number of triples$d,x,y\in[1,n]$with$xd,yd\le n$. Since there are$\sum_{k=1}^n \tau(k)$such triples with$x=y$, splitting the sum into two part according to whether$x\ge y$or$y\ge x$, we can write it as $$S = 2 \sum_{dx\le n} x - \sum_{k=1}^n \tau(k).$$ Letting$k=dx$, we get $$S = 2 \sum_{k=1}^n \sigma(k) - \sum_{k=1}^n \tau(k),$$where$\sigma\$ is the sum-of-divisors function. This gives you an identity containing arithmetic functions'', as you requested.

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