Approximation of $\frac1x\sum_{k=0}^n\frac{e^{-k^2x}-e^{-(k+1)^2x}}{2k+1}$

Question

Consider the following function $$ f(x,n)=\frac1x\sum_{k=0}^n\frac{e^{-k^2x}-e^{-(k+1)^2x}}{2k+1} $$ For small $x$, is there a way of approximating $f(x,n)$ by something like an inverse power law on $x$, depending on $n$. I am looking for something in the form $f(x,n)\simeq \alpha(n) x^{\beta(n)}$. If not, is there any function that approximates $f$, simplifying the expression on the RHS?

Note: For multiple various of $n$, for example, here are the plots of the family of functions defined by $f(x,n)$

Also, the limit $n\to\infty$ was studied for a similar function here ($xf(x,n)$).

My attempt: We can rewrite the summand as $$ \frac{e^{-k^2x}(1-e^{-(2k+1)x})}{2k+1} $$ Now, we can use the Taylor expansion of the exponential function $e^x \approx 1 + x$ for small $x$. We then get $$ 1-e^{-(2k+1)x}\approx (2k+1)x. $$ Thus, for small $x$, the summand becomes $$ \frac{e^{-k^2x}(2k+1)x}{2k+1}\approx e^{-k^2x}x $$ Using this approximation, we can rewrite the original function as $$ f(x,n)\approx \frac1x \sum_{k=0}^n e^{-k^2x}x=\sum_{k=0}^n e^{-k^2x} $$ This expression is simpler than the original one, although it's not in the form of an inverse power law. Any ideas on how to work it further?

Answer 1

The approximation of the content of the sum divided by $x$ provides $$ \frac1x\frac{e^{-k^2x}-e^{-(k+1)^2x}}{2k+1}\approx 1-\frac{1}{2}(2k^2+2k+1)x $$ Then, summing up from $0$ to $n$ gives $$ (n+1)-\frac{1}{6} \left(2 n^3+6 n^2+7 n+3\right) x $$

Answer 2

The second sum you provided, i.e. $S_n = \sum_{k=0}^n \exp(-k^2x)$, is possible to take a step further using Euler Maclaurin summation. This gives

$$ \sum_{k=0}^n f(k) \sim \int_0^n f(z)dz + \frac{1}{2}f(0) + \frac{1}{2}f(n) + \frac{B_1}{2!} \left(f^{(1)}(n) - f^{(1)}(0) \right) $$

with $f(z) = \exp(-z^2x)$ this gives that

$$ \int_0^n \exp(-z^2x)dz = \frac{\sqrt{\pi} \text{erf} (\sqrt{x}n)} {2\sqrt{x}} $$

and so one arrives at the approximation

$$ \sum_{k=0}^n \exp(-k^2 x) \sim \frac{\sqrt{\pi} \text{erf} (\sqrt{x}n)} {2\sqrt{x}} + \frac{1}{2} + \frac{1}{2}\exp(-n^2x) - \frac{1}{6}nx\exp(-n^2x) $$

E.g. setting $x=0.1$ gives that the largest relative error is around $8\cdot 10^{-5}$ and occurs for $n=1$.

This works decently for rather small $x$ for the original sum as well but even for $x=0.01$ or so the (largest) absolute error is around $6\%$ (the approximation has the same shape as the sum evaluated numerically but converges towards a somewhat lower value when $n$ increases). It would be necessary to include additional terms in the Taylor expansion and proceed likewise to account for this.