1부터 n까지 자연수의 약수의 합의 합은 다음처럼 구할 수 있다.
| 1 | 2 | 3 | 4 | 5 | 6 | … | 합 |
|---|
| 1 | 1 | 1 | 1 | 1 | 1 | … | \(1 \cdot \lfloor \dfrac{n}{1} \rfloor\) |
| | 2 | | 2 | | 2 | … | \(2 \cdot \lfloor \dfrac{n}{2} \rfloor\) |
| | | 3 | | | 3 | … | \(3 \cdot \lfloor \dfrac{n}{3} \rfloor\) |
| | | | 4 | | | … | \(4 \cdot \lfloor \dfrac{n}{4} \rfloor\) |
| | | | | 5 | | … | \(5 \cdot \lfloor \dfrac{n}{5} \rfloor\) |
| | | | | | 6 | … | \(6 \cdot \lfloor \dfrac{n}{6} \rfloor\) |
| | | | | | | | … |
| | | | | | | | \(\sum\limits_{k=1}^{n}k \cdot \lfloor \dfrac{n}{k} \rfloor\) |
\[\sum\limits_{k=1}^{n} k \cdot \lfloor \frac{n}{k} \rfloor \ {\underset{n \to \infty}\approx} \ \int_1^n k \cdot \lfloor \frac{n}{k} \rfloor\ dk\] \[t = \frac k n\] \[\int_1^n k\cdot\lfloor\frac{n}{k}\rfloor\ dk = \int_\frac{1}{n}^1 n^2 \cdot t \cdot \lfloor\frac{1}{t}\rfloor\ dt = n^2 \int_\frac{1}{n}^1 t \cdot \lfloor\frac{1}{t}\rfloor\ dt\] \[p(t) = t \cdot\lfloor\frac{1}{t}\rfloor = \left\{ \begin{array}{ll} t \quad (1 \geq t > \frac{1}{2}) \\ 2t \quad (\frac{1}{2} \geq t > \frac{1}{3}) \\ 3t \quad (\frac{1}{2} \geq t > \frac{1}{3}) \\ \vdots \\ (n-1)t \quad (\frac{1}{n-1} \geq t > \frac{1}{n}) \\ \end{array} (1 \geq t > \frac{1}{n}) \\ \right.\] \[\int_\frac{1}{n}^1 p(t)\ dt = \int_\frac{1}{2}^1 t\ dt + \int_\frac{1}{3}^\frac{1}{2} 2t\ dt +\ ... + \int_\frac{1}{n}^\frac{1}{n-1} (n-1)t\ dt\] \[= \frac{1}{2}\cdot\{(1-\frac{1}{2})(1+\frac{1}{2}) + (\frac{1}{2}-\frac{1}{3})(1+\frac{2}{3}) +\ ... + (\frac{1}{n-1}-\frac{1}{n})(1+\frac{n-1}{n})\}\] \[= \sum\limits_{a=1}^{n-1} \frac{1}{2}(\frac{1}{a}-\frac{1}{a+1})(1+\frac{a}{a+1})\] \[= \sum\limits_{a=1}^{n-1} \frac{2a+1}{2a(a+1)^2} = \sum\limits_{a=1}^{n-1} \frac{a+(a+1)}{2a(a+1)^2}\] \[= \sum\limits_{a=2}^{n} \frac{1}{2a^2} + \sum\limits_{a=1}^{n-1} \frac{1}{2a(a+1)}\]
\[\lim_{n \to \infty} (\sum\limits_{a=2}^{n} \frac{1}{2a^2} + \sum\limits_{a=1}^{n-1} \frac{1}{2a(a+1)})\] \[= \frac{1}{2} (\sum\limits_{a=1}^{\infty} \frac{1}{a^2} - 1 + \sum\limits_{a=1}^{\infty} \frac{1}{a(a+1)})\] \[= \frac{1}{2} (\frac{\pi^2}{6} - 1 + 1) = \frac{\pi^2}{12}\]
\[\sum\limits_{k=1}^{n} k \cdot \lfloor \frac{n}{k} \rfloor \approx \frac{\pi^2}{12} \cdot n^2\]