$\sum \frac{\sigma}{n}$的计数论证及$\sum_{n \leq x } \frac {\sigma(n)}{n}$渐近公式证明

Let's walk through this counting argument step by step to prove the asymptotic formula:

Step 1: Expand using the definition of $\sigma(n)$ #

First, recall that $\sigma(n)$ is the sum of all positive divisors of $n$, so:
$$\sigma(n) = \sum_{d \mid n} d$$

Substitute this into our original sum:
$$\sum_{n \leq x} \frac{\sigma(n)}{n} = \sum_{n \leq x} \frac{1}{n} \sum_{d \mid n} d$$

Step 2: Switch the order of summation #

Instead of iterating over $n$ first, we can iterate over divisors $d$ first. For each $d$, consider all multiples $n \leq x$ of $d$. This rewrites the sum as:
$$\sum_{d \leq x} d \sum_{\substack{n \leq x \ d \mid n}} \frac{1}{n}$$

Let $n = dk$ (where $k$ is a positive integer). Then $n \leq x$ implies $k \leq \frac{x}{d}$, and $\frac{d}{n} = \frac{d}{dk} = \frac{1}{k}$. Substituting this in, our sum becomes:
$$\sum_{d \leq x} \sum_{k \leq \frac{x}{d}} \frac{1}{k}$$

Step 3: Swap summation order again #

Now let's switch to iterating over $k$ first. For each $k$, $d$ can range from 1 to $\lfloor \frac{x}{k} \rfloor$. This simplifies the sum to:
$$\sum_{k \leq x} \frac{1}{k} \cdot \lfloor \frac{x}{k} \rfloor$$

Step 4: Approximate the floor function #

We know that $\lfloor \frac{x}{k} \rfloor = \frac{x}{k} + O(1)$ (the $O(1)$ term is bounded by 1). Substitute this into the sum:
$$\sum_{k \leq x} \frac{1}{k} \left( \frac{x}{k} + O(1) \right) = x \sum_{k \leq x} \frac{1}{k^2} + O\left( \sum_{k \leq x} \frac{1}{k} \right)$$

Step 5: Evaluate the asymptotic sums #

• For the first sum: The infinite series $\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$ (a classic result). The partial sum has a remainder term of $O\left( \frac{1}{x} \right)$, so:
  $$\sum_{k \leq x} \frac{1}{k^2} = \frac{\pi^2}{6} + O\left( \frac{1}{x} \right)$$
• For the second sum: The harmonic series $\sum_{k \leq x} \frac{1}{k}$ grows like $\log x + \gamma + O\left( \frac{1}{x} \right)$ (where $\gamma$ is the Euler-Mascheroni constant), so it's $O(\log x)$.

Step 6: Combine the results #

Substitute these back into our expression:
$$x \left( \frac{\pi^2}{6} + O\left( \frac{1}{x} \right) \right) + O(\log x) = \frac{\pi^2}{6}x + O(1) + O(\log x)$$

Since $\log x$ dominates the constant term, we simplify the error term to $O(\log x)$, giving the final result:
$$\sum_{n \leq x} \frac{\sigma(n)}{n} = \frac{\pi^2}{6}x + O(\log x)$$

内容的提问来源于stack exchange，提问作者Faust