It is known that the sum of the squared inverses of the divisors satisfies $$\sum _{n \leq x} \sigma_{-2}(n) = \zeta(3)x + \mathcal O(1).$$ On the other hand, an alternate calculation gives me another answer:
$$\begin{align*} \sum _{n \leq x} \sigma_{-2}(n) &= \sum _{a \leq x} \sum _{b \leq \frac xa} \frac 1{b^2}\\ & = \sum _{a \leq x} \left( -\frac ax + \zeta (2) + \mathcal O(\frac {a^2} {x^2}) \right) \\ &= -\frac 1x \left(\frac 12 x^2+ \mathcal O (x)\right) + \zeta(2)x + \frac 1 {x^2}\mathcal O (\sum _{a \leq x }a^2)\\ &= -\frac 12 x + \mathcal O(1) +\zeta(2) x + \frac 13 x + \mathcal O(1). \end{align*}$$
And this is a contradiction. I cannot spot the error in my calculation. Where did it go wrong?
Proof of known asymptotic
$$\begin{align*} \sum _{n \leq x} \sigma_{-2}(n) &= \sum _{a \leq x} \frac 1{a^2} \sum _{b \leq \frac xa} 1\\ & = \sum _{a \leq x} \frac 1 {a^2} \left(\frac xa + \mathcal O(1)\right) \\ &= x \sum_{a \leq x }\frac 1 {a^3} + \mathcal O(\sum _{a \leq x} \frac 1 {a^2}) \\&= \zeta(3)x + \mathcal O(1). \end{align*}$$
You are tacitly assuming that a constant is $1$, while that is not the case.
$$ \sum_{b\leq N}\frac{1}{b^2} = \zeta(2)-\frac{1}{N}+\frac{1}{2N^2}-\frac{1}{6N^3}+\frac{1}{30N^5}+\ldots $$ hence by replacing $N$ with $\frac{x}{a}$ we get $$ \sum_{b\leq\frac{x}{a}}\frac{1}{b^2} = \zeta(2)-\frac{a}{x}+\frac{a^2}{2x^2}-\frac{a^3}{6x^3}+\frac{a^5}{30x^5}+\ldots $$ and summing both sides over $a\leq x$ we have
$$ \sum_{a\leq x}\sum_{b\leq\frac{x}{a}}\frac{1}{b^2} = x\left(\zeta(2)-\frac{1}{2}+\frac{1}{6}-\frac{1}{24}+\frac{1}{180}+\ldots\right)+\mathcal{O}(1) $$ where by the Euler-Maclaurin summation formula the involved constant is exactly $\zeta(3)$ and not $\zeta(2)-\frac{1}{6}$.