Is it possible to express $\sqrt2$ as a convergent infinite summation using only integer arithmetic and nested finite sums?

Question

I’m exploring a restricted arithmetic framework and have arrived at the following conjecture:

Tyler’s 2025 Conjecture:
There exists no infinite summation of the form $$\sum_{k=0}^{\infty} f(k)$$ that converges exactly to $\sqrt{2}$, where $f(k)$ is constructed only from:

•   Integer constants and the summation index k,
•   A finite number of additions, subtractions, multiplications, integer divisions, and integer exponentiations (including negative exponents),
•   Optionally, finite nested summations (also constrained to this grammar),

and excluding:

•   Non-integer exponents (e.g. fractional powers),
•   Factorials or binomial coefficients,
•   Transcendental functions (e.g. sin, exp, log),
•   Recursion, externally defined constants, or products over a range.

Extended Rule:

Constants like $\pi, e,$ or $\ln (2)$ may appear only if they are themselves expressed within the same framework, i.e., as legal summations using this grammar.

Why I believe it’s true:

•   Every $f(k)$ in this framework yields rational values, so each partial sum is rational.
•   The only known summations that converge to $\sqrt{2}$ involve forbidden constructs like radicals, fractional powers, binomial coefficients, or products.
•   Approximating $\sqrt{2}$ arbitrarily closely using only rational partial sums with “too-good” convergence runs afoul of Roth’s Theorem.
•   Even allowing legal definitions of $\pi$ or $e$ doesn’t help: no known rational transformation from them reaches $\sqrt{2}$ under these constraints.

This feels like a hard barrier: even though $\sqrt{2}$ is computable and algebraic, it may be inaccessible to this kind of purely integer-based summation logic.

⸻

My questions:

1.  Has anything like this conjecture been proven or refuted in number theory or computable analysis?
2.  Can we rigorously characterize the set of numbers representable by such constrained summations?
3.  Is there a known summation (using only this grammar) that converges to $\sqrt{2}$ or provably cannot?

I’m aware this touches on computability, irrational approximation, and perhaps symbolic logic. Any direction, refutation, or constructive idea would be appreciated.

Answer 1

I think I can do this:

$eq(x,y)=\sum_0^{-(x-y)^2}(1)$

Note $eq(x,y)=1$ when $x=y$ and $eq(x,y)=0$ when $x\ne y$.

We have $\left\lfloor\frac xy\right\rfloor=\sum_{a=1}^x\sum_{b=1}^x eq(ay,b)$ whenever $x$ is a non-negative integer and $y$ is a positive integer.

Now we get $x\mod y=x-\left\lfloor\frac xy\right\rfloor\times y$ whenever $x$ is a non-negative integer and $y$ is a positive integer.

Then we can get $\binom xy=\left(\left\lfloor\frac{(1+2^x)^x}{2^{xy}}\right\rfloor\mod 2^x\right)+eq(x+y,0)$ for non-negative integers $x$ and $y$.

Finally, $\sqrt2=\sum_{k=0}^\infty\binom{2k}{k}\frac1{8^k}$.

Answer 2

We have $$ \sqrt{2} = \frac{\sum\limits_{n=0}^\infty \frac{ 4(-\frac12)^n + (-\frac18)^n}{2n+1}}{\sum\limits_{n=0}^\infty 4 \frac{(-1)^n}{2n+1}} $$

I'm drawing on formulas from this paper by Bailey, Borwein, and Plouffe. Their paper claims it's an open problem whether there is an integer number $b$ and integer-coefficient polynomials $p$ and $q$ such that: $$ \sqrt{2} = \sum_{n=0}^\infty b^{-n} \frac{p(n)}{q(n)} $$

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Proof of formula: It follows immediately from these identities: \begin{eqnarray} \pi &=& \sum_{n=0}^\infty 4 \frac{(-1)^n}{2n+1}\\ \pi \sqrt{2} &=& \sum_{n=0}^\infty \frac{ 4(-\frac12)^n + (-\frac18)^n}{2n+1} \end{eqnarray}

The first of these is an evaluation of $4\arctan(1)$ by Taylor series. The second is equation 2.13 in Bailey, Borwein, and Plouffe. It is proven as follows:

By the angle addition formula for arctangents, \begin{eqnarray} 2\arctan \frac 1{\sqrt 2} &=& \arctan\left(\frac{\frac{1}{\sqrt2}+\frac1{\sqrt2}}{1-\frac1{\sqrt{2}\cdot\sqrt 2}}\right)\\ &=& \arctan\left(2\sqrt2\right) = \arctan \sqrt 8 \end{eqnarray} Then, using the fact that $\arctan(x) + \arctan(1/x) = \frac\pi2$ for all $x$, we have: $$ 2\arctan \frac1{\sqrt 2} + \arctan\frac1{\sqrt 8} = \frac\pi 2 $$ Using the Taylor series of arctan:\begin{eqnarray} 2\arctan \frac1{\sqrt 2} + \arctan\frac1{\sqrt 8} &=& 2\sum_{n=0}^\infty (-1)^n \frac{(\frac1{\sqrt 2})^{2n+1}}{2n+1}+\sum_{n=0}^\infty (-1)^n \frac{(\frac1{\sqrt 8})^{2n+1}}{2n+1}\\ &=& 2\frac1{\sqrt2}\sum_{n=0}^\infty (-1)^n \frac{(\frac1{\sqrt 2})^{2n}}{2n+1}+\frac1{\sqrt8}\sum_{n=0}^\infty (-1)^n \frac{(\frac1{\sqrt 8})^{2n}}{2n+1}\\ &=& \frac1{2\sqrt2}\sum_{n=0}^\infty (-1)^n \frac{4(\frac12)^{n}}{2n+1}+\frac1{2\sqrt2}\sum_{n=0}^\infty (-1)^n \frac{(\frac1{ 8})^{n}}{2n+1} \\ &=& \frac1{2\sqrt 2}\left(\sum_{n=0}^\infty (-1)^n\frac{4(\frac12)^{n}+(\frac1{ 8})^{n}}{2n+1}\right) \end{eqnarray}

Then multiply both sides by $2\sqrt2$ to obtain the series for $\sqrt 2 \pi$.

Answer 3

Using the Taylor series for $f(t)=(1-t)^{1/2}$ you can see that $$ \sqrt2=1-\sum_{n=1}^\infty \frac{(-1)^n2(2n-2)!}{4^nn!(n-1)!}. $$