Assume the following sum: $$ s(x) = \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty}\frac{1}{n \sqrt{n^2+x^2} \left(\sqrt{n^2+x^2}+n\right)}, $$ where $x \in \mathbb{R}$. Since $f_n(x) = f_n(-x)$, it suffices to consider $x \geq 0$. The functions $f_n$ are bounded by $$ |f_n(x)| \leq \frac{1}{2 n^3} ~ \forall x \geq 0, $$ this implies that $f_n(x) \rightrightarrows 0$. Because $\sum n^{-3}$ converges, it also follows that $\sum f_n(x) \rightrightarrows s(x)$ absolutely for $x \geq 0$. Since there is no closed formula for the sum, I am interested in series of expansion. At $x=0$ it works well, but I am struggling with series at $x=\infty$ . The naive approach gives $$ f_n(x) = \frac{1}{n x^2} -\frac{1}{x^3} + \frac{n^2}{2 x^5}+O\left(\frac{1}{x^7}\right), $$ which cannot be used as the first leading term goes as $1/n$ and it diverges in the sum (as well as the other terms). If I let Mathematica compute the plot, it looks like this:
Of course, if I transform to $y=1/x$, then $$ f_n(y) = \frac{y^2}{n \left(n^2 y^2+n y \sqrt{n^2 y^2+1}+1\right)} $$ and the data near $y = 0$ can be fitted, e.g.
or
but there are many fits which can fit well. I am looking for expansion of $s(x)$ for $1/x \ll 1$. Is there an analytical way how to obtain the series? Is my approach correct?
Thank you for any advice.
Edit: for example, series of $$ \tanh(x) = - 1 + \frac{2}{1+e^{-2x}} $$ for $1/x \ll 1$ is not polynomial in $1/x$, but it can be done in terms of $y=e^{-2x}$ at $y=0$, then $$ \tanh(x) = 1-2 e^{-2x}+2 e^{-4x}+O\left(e^{-6x}\right). $$ Ex. 2: this function has expansion (Laurent series) at $x = \infty$ in powers of $1/x$: $$ \frac{1}{1+x^2} = \frac{1}{x^2} - \frac{1}{x^4} + O\left(\frac{1}{x^6}\right). $$