Average abundance of squares

Lemma:

Let $\displaystyle G(s) = \sum_{n=1}^{\infty} \frac{F(n)}{n^s}$ and $F = 1 \star f$ , where $\star$ is the Dirichlet Convolution . Then $\displaystyle \lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n\le N}F(n) = \lim_{s\rightarrow 1}\frac{G(s)}{\zeta(s)}$

Proof: $\begin{aligned} \lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n\le N} F(n) &= \lim_{N\rightarrow \infty} \frac{1}{N}\sum_{d\le N}f(d)\sum_{d|n,n\le N}1 \\ &= \lim_{N\rightarrow \infty} \frac{1}{N} \sum_{d\le N} f(d) \left\lfloor \frac{N}{d} \right\rfloor \\ &= \lim_{N\rightarrow \infty} \sum_{d \le N} \frac{f(d)}{d} \\ &= \lim_{s \rightarrow 1} \frac{G(s)}{\zeta(s)} \end{aligned}$

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Now, using $\displaystyle \sum_{n = 0}^{\infty} \frac{F(n)}{n^s} = \prod_{p} \sum_{a=0}^{\infty} \frac{F(p^a)}{p^{as}}$ , if $F$ is a multiplicative function,

$\sum_{n = 0}^{\infty} \frac{\sigma(n^k)}{n^s} = \prod_p \sum_{a=0}^{\infty} \frac{1-p^{ak+1}}{1-p} = \prod_p \frac{p^s(p^k+p^{s+1}-p^s-p)}{(p-1)(p^s-1)(p^s-p^k)}$

Now let $\displaystyle F(n) = \frac{\sigma(n^k)}{n^k}$ , then $\begin{aligned} \displaystyle G(s) &= \sum_{n=1}^{\infty} \frac{\sigma(n^k)}{n^{s+k}} \\ \displaystyle &= \prod_p \frac{p^{k+s}(p^{k}+p^{k+s+1}-p^{k+s}-p)}{(p-1)(p^{k+s}-1)(p^{k+s}-p^{k})} \\ \displaystyle &= \prod_p \frac{\left(\frac{1}{p^{s+1}}-\frac{1}{p^{s+k}}-\frac{1}{p}+1\right)}{\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^{s+k}}\right)\left(1-\frac{1}{p^{s}}\right)} \end{aligned}$

Via the lemma, $\displaystyle \lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n\le N} F(n) = \lim_{s \rightarrow 1} \prod_p \frac{\left(\frac{1}{p^{s+1}}-\frac{1}{p^{s+k}}-\frac{1}{p}+1\right)}{\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^{s+k}}\right)}$

For $k=2$ , this can be expressed as

$\lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n\le N} F(n) = \prod_p \frac{\left(1-\frac{1}{p}+\frac{1}{p^{2}}-\frac{1}{p^{3}}\right)}{\left(1-\frac{1}{p}\right)\left(1-\frac{1}{p^{3}}\right)} = \frac{\left(1-\frac{1}{p^{4}}\right)}{\left(1-\frac{1}{p^{2}}\right)\left(1-\frac{1}{p^{3}}\right)} = \frac{\zeta(2)\zeta(3)}{\zeta(4)}$

Similarly, for $k \rightarrow \infty$ , $\displaystyle \lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n\le N} F(n) = \frac{\zeta(2)\zeta(3)}{\zeta(6)}$ .

As an exercise, try to prove that for all integer $3 \le k$ , $\displaystyle \lim_{N\rightarrow \infty} \frac{1}{N}\sum_{n\le N} F(n)$ cannot be expressed as a product of $\zeta(i)^j$ where $i \ge 2$ and $j$ are integers.