I'm looking for an explicit fomula for calculating the natural number $n$, as result value, from any (finite) given argument value $$\text{Sum} \left[ \frac{1}{2 k + 1}; \{ k, 0, n \} \right].$$
Hopefully such a formula would be explicit and short enough to be entered and plotted through the Wolfram Alpha website interface; and ideally it provides some monotonous real-valued interpolation between successive integers $n$ and $(n + 1)$ for arbitrary positive real argument values $x$ with
$$\text{Sum}\left[ \frac{1}{2 k + 1} ; \{ k, 0, n \} \right] \lt x \lt \text {Sum} \left[ \frac{1}{2 k + 1} ; \{ k, 0, n + 1\} \right].$$
Note (hint?):
Web-searching for "summing receprocials of consecutive odd integers" I've come across a relation to the so-called "Jeep problem", and from this treatment I gather (and checked for small values $n$) that the explicit formula
$$\frac{\text{EulerGamma}}{2} + \text{Log}[ \, 2 \, ] + \frac{\text{PolyGamma}[ \, n + 0.5 \, ]}{2} \tag{*}$$
is suitable for calculating the result value $\text{Sum} \left[ \frac{1}{2 k + 1} ; \{ k, 0, n \} \right]$ from natural arguments $n$.
(I'm not even sure whether that's correct for arbitrary large arguments $n$; so I'd appreciate a proof of that, or any counter-example, too.)
If so, my main question, again, is for some (hopefully not too complicated) expression of the inverse of formula $(*)$.
$$S_n=\sum\limits_{k=0}^n \frac{1}{2 k+1}=\frac{1}{2}\, H_{n+\frac{1}{2}}+\log (2)$$ If $n$ is large $$S_n=\left(\log(2)+\frac \gamma 2\right)+\frac 12\log(n)+\frac{1}{2 n}-\frac{11}{48 n^2}+O\left(\frac{1}{n^3}\right)$$ Ignoring the last term $$S_n=\left(\log(2)+\frac \gamma 2\right)+\frac 12\log(n)+\frac{1}{2 n}$$ leads to $$n=-\frac{1}{W\left(-4\, e^{\gamma -2 S_n}\right)}$$ where $W(.)$ is Lambert function.
This is a quite good approximation; for $n=5$, it will return $n=4.8975$. So, use $$\large\color{blue}{n=\left\lceil -\frac{1}{W\left(-4\, e^{\gamma -2 S_n}\right)} \right \rceil}$$
Edit
If you want an approximation (before rounding), let $$k=4e^{\gamma} \qquad \qquad t=k\,e^{-2S_n}$$ Using series $$n=\frac 1 t- \sum_{m=0}^\infty a_m\,t^m$$ where the first coefficients are $$\left\{1,\frac{1}{2},\frac{2}{3},\frac{9}{8},\frac{32}{15},\frac{ 625}{144},\frac{324}{35},\frac{117649}{5760},\frac{131072}{283 5},\frac{4782969}{44800}\right\}$$
The numerators and denominator form respectively sequences $A264234$ and $A264235$ in $OEIS$.
Transforming the summation into a Padé approximant
$$n=\frac 1 t -\frac{665 t^2-2226 t+1020}{1353 t^2-2736 t+1020}$$ whose error is $\sim \frac{11 }{100}t^5$.