In my first year of high school, I was given a chemistry problem with roots in pure math (it had to do with equilibrium, but the details aren't important here.) A common tactic in complicated mass-charge balance questions of that caliber is to express all unknown concentrations as an expression of one, then get an equation of the form $x=f(x).$ You then enter $f(Ans)$ into your calculator, make an initial guess, and then press enter until the answer converges.
Since this problem was much more math than chemistry, though, you actually end up with an infinite sum! (The original phrasing of the problem involved adding infinite acids to a bottle.) The equation I arrived at was this: $$[H^+]=\sum^{\infty}_{i=1}\frac{1\times 10^{-6}\cdot 0.5^i}{i[H^+]+1\times 10^{-6}}.$$
I tested the value of the RHS for several guesses of $[H^+]$ using the first 10 terms (notably, this was on a TI calculator with 2 lines of LCD display.) Most telling was a sum of $6.93 \times 10^{-4}$ for $[H^+]=1 \times 10^{-3}.$ Thus, I conjectured that the RHS simplified to $\ln{(2)}\cdot 1\times 10^{-6}/[H^+].$ From there, solving the equation was trivial (no iteration needed, thank god.) Then, substituting the result of $8.33 \times 10^{-4}$ back into the expression gave the same number back, confirming the conjecture.
My question is, how would you figure out that summation through math and not bs approximations? I'm noting a superficial similarity to the power series expansion of $\ln{x}$ but can't quite work it out. An answer would be greatly appreciated to this problem which has been living rent-free in my head for a while.
Changing notations $$x=[H^+] \qquad k=10^{-6}$$ you look for the zero of the equation $$x=k\sum_{i=1}^\infty \frac{2^{-i}}{i x+k}=\frac k{2x}\,\Phi \left(\frac{1}{2},1,1+\frac{k}{x}\right)$$ where $\Phi(.)$ is the Hurwitz-Lerch transcendent function.
Let $x=\frac k y$ and find the zero of function $$f(y)=y^2 \,\,\Phi \left(\frac{1}{2},1,y+1\right)-2 k$$
Expanding around $y=0$,we then have $$f(y)=-2k+2\log (2)y^2+\frac{6 \log ^2(2)-\pi ^2}{6}y^3+O\left(y^4\right)$$ which gives as a first estimate $$y_0=\frac{\sqrt{k}}{\sqrt{\log (2)}} \implies x_0=\sqrt{k\log (2)}=8.32555\times 10^{-4}$$
The exact solution is $$y=1.20173\times 10^{-3} \implies x=8.32135\times 10^{-4}$$
Edit
We could make it better using, instead of Taylor series, the simple $[2,2]$ Padé approximant and solve the quadratic given by $$\frac {2 y^2 \log (2) } {1+\frac{ \left(\pi ^2-6 \log ^2(2)\right)}{12 \log (2)}y+\frac{ \left(-126 \zeta (3) \log (2)+\pi ^4+12 \log ^4(2)\right)}{144\log ^2(2)}y^2 }=2k$$ which gives $$y=1.20173\times 10^{-3}$$ which is in an absolute error of $3.875\times 10^{-14}$.
Update
Notice also that, if we neglect the $k$ in denominator (this is justified by the fact that you are mixing a bunch of acids), the problem is to solve $$x=\frac{k \log (2)}{x} \implies x_0\sim \sqrt{k\log (2)}$$ which was the very first estimate.
Now, one iteration of Newton method and series expansion for small $k$ leads to $$x_1=t\left(1+ \left(\frac{1}{4}-\frac{\pi ^2}{24 \log ^2(2)}\right)\,t+O\left(t^2\right) \right)\qquad \text{with}\qquad t=\sqrt{k\log (2)}$$
Truncated at this level, this gives numerically $$x_1=8.32135\times 10^{-4}$$ which is in a relative error of $1.774\times 10^{-7}$ % $\large (!!)$.