Proving a rational number is NOT in the lower Dedekind cut for a transcendental number I'm attempting to argue with a finitist who claims that transcendental numbers can't be defined as Dedekind cuts without using an infinite predicate or in some other way requiring an infinite number of steps.
Since Euler proved that $\sum_{n=0}^\infty \frac{1}{n^2}$ converges to $\frac{\pi^2}{6}$ my approach is to define the lower cut like this:
$$\pi = \left\{ a \in \Bbb Q | (a \lt 0) \lor \exists n \in \Bbb N: a^2 < \sum_{i=1}^n\frac{6}{i^2} \right\}$$
This works quite well for showing that some rational number, such as $\frac{5}{2}$, belongs in the lower cut.  I can pick n = 2 and show that $\left(\frac{5}{2}\right)^2 < \frac{6}{1^2} + \frac{6}{2^2}$ because $\frac{25}{4} < \frac{24 + 6}{4}$.  Easy!  The same approach works with larger rational numbers as long as I'm proving that they DO belong in the cut.  However, I'm a bit lost when trying to prove that something like $\frac{7}{2}$ does NOT belong in the cut.  Obviously, I could just say that the digits of $\pi$ are 3.14159..., so 3.5 is not less than that, but I think my argument would be stronger if I could show a proper proof using Euler's series.
This is essentially similar to the claim that $\sum_{n=0}^\infty \frac{1}{2^n}$ never reaches 3.  Obviously it doesn't, since it converges on 2.  But how does one prove that in a finite number of steps?  By "a finite number of steps" here I mean that I'm trying to avoid relying on the fact that the series converges, since that opens me up to the charge that converging takes an infinite number of steps.  I'm looking for a method like the inclusion proof where some finite number of steps always suffices.  Is that possible?
 A: To show that $7/2 > \pi$, it suffices to check that $(7/2)^2 > \pi^2$. We follow a simple process.

*

*Approximate $\pi^2$ to within $\epsilon$ - let $\kappa$ be the approximation.

*Check to see if $(7/2)^2$ is within $\epsilon$ of $\kappa$. If not, then we know whether $7/2 > \pi$ or $7/2 < \pi$. If so, go back to step 1 but reduce $\epsilon$ to $\epsilon / 2$.

In this case, I will approximate $\pi^2$ to within $0.3$. To do this, I need to find some $n$ such that $|\pi^2 - \sum\limits_{i = 1}^n \frac{6}{i^2}| < 0.3$. I can do this by using some proof of convergence of the series. In fact, it turns out that we can use $n = 20$ since $\pi^2 - \sum\limits_{i = 1}^n \frac{6}{i^2} = \sum\limits_{i = n + 1}^\infty \frac{6}{i^2} < \int\limits_{i = n}^\infty \frac{6}{i^2} = \frac{6}{n} = 0.3$. The approximation is $\kappa = \sum\limits_{j = 1}^{20} \frac{6}{j^2} = \frac{17299975731542641}{1806412533045120}$. And it's easy to see that $\kappa + 0.3 = \frac{17841899491456177}{1806412533045120} < (7/2)^2$. Therefore, we have $\pi^2 \leq \kappa + 0.3 < (7/2)^2$. So $\pi < 7/2$.
Note that this can easily be translated into a proof that $7/2$ is not in the given Dedekind cut. Although OP slightly misdefined the cut - it should be $\{a \in \mathbb{Q} | a < 0 \lor \exists n \in \mathbb{N} (a^2 < \sum\limits_{i = 1}^n \frac{6}{i^2})\}$.
