Finding $\lim_{n\to\infty}\frac1{n}\sum_{k=1}^n(\alpha(k)-\beta(k))$, where $\alpha$ and $\beta$ count even and odd divisors, respectively

Question

For every $n\in \mathbb{N}^{*}$, let $\alpha(n)$ be the number of even divisors of n and $\beta(n)$ be the number of odd divisors of n. Calculate:

$$\lim_{n\to\infty} \frac{1}{n}\cdot \sum_{k=1}^{n}(\alpha(k)-\beta(k))$$

Answer 1

Denote the sums of the $\alpha(k)$ and $\beta(k)$ by $A(n)$ and $B(n)$ respectively.

When counting the odd divisors of $1,2,..., n$ we are counting the number of multiples of $1,3,5,...$ which are no greater than $n$ and so

$$B(n)= n+ \left \lfloor{\frac{n}{3}}\right \rfloor+\left \lfloor{\frac{n}{5}}\right \rfloor+...$$ Similarly, $$A(n)= \left \lfloor{\frac{n}{2}}\right \rfloor+\left \lfloor{\frac{n}{4}}\right \rfloor+\left \lfloor{\frac{n}{6}}\right \rfloor+...$$

Thus $$\frac{B(n)-A(n)}{n}\approx\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+... $$