In 2011 when I was in school I created a formula to calculate square roots... For $x\in\mathbb{R}$ with $x>0$ the following holds:
$$\sqrt{x} = \sum\_{n=0}^{\infty}\frac{\left(\prod\_{k=1}^{n}\left(\frac{3}{2}-k\right)\right)\left(\frac{1}{4}+\frac{x}{2}+\frac{x^2}{4}\right)^{\frac{1}{2}-n}\left(-\frac{1}{4}+\frac{x}{2}-\frac{x^2}{4}\right)^{n}}{n!}$$


It was mostly a joke and the formula is not useful. Recently I stumbled upon this formula again, and decided to work on the ideas behind that formula, not on the formula itself, to create new formulas that are actually useful. With "useful" I mainly mean a good rate of convergence, whatever "good" exactly means in this context...


If you're curious about the specifics of how everything unfolded (and if you either understand German or are willing to use a translator): [The regarding thread on matheplanet.com](https://matheplanet.com/matheplanet/nuke/html/viewtopic.php?topic=162777)


I've finished my work for now and, in my opinion, the results are quite nice. However, as I don't have a background in mathematics (I did not study mathematics), I don't know whether these things already exist, or whether they could potentially be useful. I haven't proven anything yet, but I am still confident that it's all correct.


I'd be grateful to get some information about my results from people who actually know mathematics, since I really have no idea what to think of my work.


Let $x,a \in \mathbb{R}\_+$ and $m \in \mathbb{N}\_0$ and let $\sqrt{x} \neq a > \sqrt{\frac{x}{2}} > 0$ hold


We define
$$S(x,a,m) := \sum\_{k=0}^{m}\left(\sum\_{n=k}^{m}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\binom{n}{k}\right)\left(-1\right)^{k+1}a^{1-2k}x^{k}\right)$$
$$S(x,a) := \lim\_{m\to\infty}S(x,a,m) = \sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}a^{1-2k}x^{k}\right)\right)$$
It holds:
$$\forall m : S(x,a,m) \neq S(x,a) = \sqrt{x}$$


We define $\hat{a} := x+1$ and $z := S(x,\hat{a},m)$ which yields:
$$\sqrt{x} = S(x,z) = S(x,S(x,x+1,m))$$


Since I don't have any proofs, I used Mathematica at least... The [Mathematica output](https://i.stack.imgur.com/eBbiM.png) and the regarding [Mathematica notebook](https://matheplanet.com/matheplanet/nuke/html/dl.php?id=2472&1683559031).


Using [Nest](https://reference.wolfram.com/language/ref/Nest.html) I created a really quick & dirty (& (very) inefficient) iteration to make it converge fast.


Iterating $5$ times with $m=10$ and $x=2$ yields over $34000$ correct decimal places for $\sqrt{2}$.


$m=11$ yields over $51000$ correct decimal places.


It boils down to: The bigger $m$, the faster the iteration converges... $\text{¯\\_(ツ)\_/¯}$


In other words: You calculate $S(x,a,m)$ for a given $m$ once, the result is an easy function, see the [Mathematica output](https://i.stack.imgur.com/eBbiM.png) for $S(x,x+1,10)$ as an example. You can save that function, and use it for the iteration.


Furthermore $S(x,a,m)$ and consequently $S(x,x+1,m)$ are (hopefully) easy to compute.
Note that the outer sum runs from $k=0$ to $m$ and $a^{1-2k}x^{k}$ is nicely isolated at the end, so that everything in front of $a^{1-2k}x^{k}$ yields the coefficients.


For $S(x,a)$ everything is "rearranged", so that we do not get nested infinity due to $m\to\infty$.


To get nice $\sqrt{x}$ formulas we can e. g. do the following...
For $x\in\mathbb{R}$ and $x>0$ holds:
$$\sqrt{x} = \sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\left(x+1\right)^{1-2k}x^{k}\right)\right) = \sum\_{n=0}^{\infty}\left(\frac{\Gamma\left(n-\frac{1}{2}\right)}{2\sqrt{\pi}n!}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\left(x+1\right)^{1-2k}x^{k}\right)\right)$$


Kind of by accident (I looked at $\sqrt{\pi}$) I also obtained the following two formulas:
$$\sqrt{\pi}=\int\_{-\infty}^{\infty}e^{-x^2}\,dx = \sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\pi^{1-k}\right)\right)$$
$$\pi = \sum\_{n=0}^{\infty}\left(\frac{\Gamma\left(n-\frac{1}{2}\right)}{2n!}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}\pi^{1-k}\right)\right)$$


The last one is just supposed to look cool... For $x \in \mathbb{R}$ and $x > 0$ and $m \in \mathbb{N}\_0$ holds:


$$\sqrt{x}=\sum\_{p=0}^{\infty}\left(-1\right)^{p-1}\frac{\left(\sum\_{q=0}^{m}\left(-1\right)^{q-1}\frac{\left(x+1\right)^{1-2q}}{q!}\frac{\Gamma\left(q-\frac{1}{2}\right)}{2\sqrt{\pi}}\left(x-(x+1)^2\right)^{q}\right)^{1-2p}}{p!}\frac{\Gamma\left(p-\frac{1}{2}\right)}{2\sqrt{\pi}}\left(x-\left(\sum\_{r=0}^{m}\left(-1\right)^{r-1}\frac{\left(x+1\right)^{1-2r}}{r!}\frac{\Gamma\left(r-\frac{1}{2}\right)}{2\sqrt{\pi}}\left(x-(x+1)^2\right)^{r}\right)^2\right)^{p}$$


**Edit**:


* I guess I should include, that I've never been able to speak to someone about my mathematics in my whole life, so I really have no idea...
* "Working in isolation" was mentioned in the comments. I'd like to clarify: Although I've worked in isolation only, that's not what I want. I simply didn't know who to speak to...


**Answer to [answer](https://mathoverflow.net/a/446587/504411)**:
While thinking about writing down what I did in regard to the Taylor series of $\sqrt{x}$ I noticed that I'm now able to ask specific questions.


However: I haven't developed abstract enough mathematics to express those questions, yet.


Consequence: I'll develop the needed mathematics, and when I've done so I'll edit this question (again) to ask those questions and I'll also explain what I did in regard to the Taylor series of $\sqrt{x}$ to give some context.


(Or should I open a new post, then? If so, please let me know)



New method to compute square roots
Let me "unclutter" the basic formula $S(x,a)=\sqrt{x}$, starting from the definition in the OP,
$$S(x,a) =\sum\_{n=0}^{\infty}\left(\frac{\left(n+1\right)\binom{2n+2}{n+1}}{\left(4n^2-1\right)2^{2n+1}}\sum\_{k=0}^{n}\left(\binom{n}{k}\left(-1\right)^{k+1}a^{1-2k}x^{k}\right)\right).$$
The finite sum over $k$ is the [binomial expansion](https://en.wikipedia.org/wiki/Binomial_theorem) of $-a(1-x/a^2)^n$, and with some further algebra we have
$$S(x,a)=-\frac{a}{2\sqrt{\pi}}\sum\_{n=0}^\infty\frac{\Gamma \left(n-\frac{1}{2}\right) }{ \Gamma (n+1)}\left(1-x/a^2\right)^n.$$
This is simply the Taylor series of $a\sqrt {x/a^2}=\sqrt{x}$ around $x=1$, independent of $a>0$.


So I am unsure whether $S(x,a)$ has more content than just this, a power series of $\sqrt x$ around $x=1$.


