Consider the infinite series $$f(z) = 2\sum_{k=0}^\infty(-1)^k z^{(2k+1)^2} -1.$$ I want to show that $f$ admits one zero in the interval $(0,1)$ (in $\mathbb{R}$).
I have perfect knowledge about the nature of zeros for $z$ away from 1, since I can approximate it by truncating the series. However, the truncation does not work well as $z\to 1^-$.
I am thinking to use techniques in complex analysis. One possibility is to apply Rouche's theorem, by comparing $f$ with function $g$ like $g(z)=2z-1$ for $z > 0.5$. However I have difficulty showing the condition $|f(z)-g(z)|<|f(z)|+|g(z)|$.
What possible approach one can attempt to say something about the nature of zeros as $z\to 1^-$?
The following figure shows the plot of the function $f$
$f(z)$" />
I also have the following guess: $f(z)<-1+2z-z^4$ for $z$ close to 1, which will help me to see the nature of roots.
If you consider the function $$f_n(z) = 2\sum_{k=0}^n(-1)^k z^{(2k+1)^2} -1$$ there is a root close to $z=1^-$ only if $n$ is odd. If $n$ is even, close to $z=1^-$ there is a positive local minimum value but no root.
To estimate the root, what you can do is to let $z=1-\epsilon$, expand $$(1-\epsilon )^{(2 k+1)^2}=\sum_{p=0}^\infty (-1)^p \binom{(2 k+1)^2}{p}\epsilon^p$$ truncate to some order, perform the summation, consider that it is a trucated series to some order and use series reversion.
For example, performing the above summation up to $p=5$ and $k=3$ leads to $$g(\epsilon)=4-168 \epsilon +3024 \epsilon ^2-41616 \epsilon ^3+449304 \epsilon ^4-3920280 \epsilon ^5+O\left(\epsilon ^{6}\right)$$ $$\epsilon=-\frac{g(\epsilon)-4}{168}+\frac{(g(\epsilon)-4)^2}{1568}-\frac{467 (g(\epsilon)-4)^3}{5531904}+\frac{66781 (g(\epsilon)-4)^4}{5576159232}-\frac{11456513 (g(\epsilon)-4)^5}{6557563256832}+O\left((g(\epsilon)-4)^6\right)$$ SInce we want $g(\epsilon)=0$, the estimate is $$\epsilon=\frac{283507943}{6403870368}=0.0443$$ while the exact solution is $\epsilon=0.0433$.
What it seems is that $$f_{2m+1}(z)=0 \quad \implies \quad z\sim 1-\frac 1 {25 m}$$ Starting with such estimate, Newton method would converge quite fast.
Trying for $m=10$ (which corresponds to a polynomial of degree $1849$), Newton iterates will be $$\left( \begin{array}{cc} q & z_q \\ 0 & 0.996000 \\ 1 & 0.998004 \\ 2 & 0.997554 \\ 3 & 0.997211 \\ 4 & 0.997049 \\ 5 & 0.997022 \\ 6 & 0.997021 \end{array} \right)$$