Say I have some function $f(n,x)$. For many "nice" functions, I find (numerically) that \begin{equation} \sum_{n=1}^\infty r_2(n) f(n,x)\approx\pi\int_0^\infty dn f(n,x). \end{equation} Here $r_2(n)$ denotes the sum of squares function. I am aware that the average value of the sum of squares function is asymptotically $\pi$, which is how I even thought of trying this.
I am ideally looking for a way to bound the error for an arbitrary function, or for some class of reasonable functions.
My intuition says we have to take into account some bounds on the derivative of the function. I tried converting the sum to 2D, but couldn't find a nice way to use the Abel-Plana formula or something related to get my bound. I also briefly tried reasoning about this measure-theoretically, but without much success.
Assume that $f(0,x) = 0$ and $\lim _{n \to + \infty } n \cdot f(n,x) = 0$ for any fixed $x$. Then, by Abel's summation formula, $$ \sum\limits_{n = 1}^\infty {r_2 (n)f(n,x)} = - \int_0^{ + \infty } {\frac{{\partial f(t,x)}}{{\partial t}}\left( {\sum\limits_{k = 1}^t {r_2 (k)} } \right){\rm d}t} . $$ Now $$ \sum\limits_{k = 1}^t {r_2 (k)} = \pi t + \mathcal{O}(t^\alpha ) $$ with some $\alpha>0$ (the best result currently is due to Huxley: $\alpha = 131/416=0.3149038461\ldots$, the conjecture is that $\alpha=1/4+\varepsilon$). Then \begin{align*} \sum\limits_{n = 1}^\infty {r_2 (n)f(n,x)} & = - \pi \int_0^{ + \infty } {\frac{{\partial f(t,x)}}{{\partial t}}t\,{\rm d}t} + \int_0^{ + \infty } {\frac{{\partial f(t,x)}}{{\partial t}}\mathcal{O}(t^\alpha) \, {\rm d}t} \\ & = \pi \int_0^{ + \infty } {f(t,x)\,{\rm d}t} + \int_0^{ + \infty } {\frac{{\partial f(t,x)}}{{\partial t}}\mathcal{O}(t^\alpha ) \,{\rm d}t}, \end{align*} provided that the improper integrals converge.
As an example, we have $$ \sum\limits_{n = 1}^\infty {\frac{{r_2 (n)}}{{x^2 + n^2 }}} = \frac{{\pi ^2 }}{{2\left| x \right|}} + \mathcal{O}\!\left( {\frac{1}{{\left| x \right|^{701/416} }}} \right) $$ as $|x|\to+\infty$.