How are the known digits of $\pi$ guaranteed?

Note that $\pi=6\arcsin\left(\frac12\right)$. So, since$$\arcsin(x)=\sum_{n=0}^\infty \frac1{2^{2n}}\binom{2n}n\frac{ x^{2n+1}}{2n+1},$$you have$$\pi=\sum_{n=0}^\infty\frac6{2^{4n+1}(2n+1)}\binom{2n}n.$$Now, for each $N\in\mathbb{Z}^+$, let$$S_N=\sum_{n=0}^N\frac6{2^{4n+1}(2n+1)}\binom{2n}n\text{ and let }R_N=\sum_{n=N+1}^\infty\frac6{2^{4n+1}(2n+1)}\binom{2n}n.$$Then:

• $(\forall N\in\mathbb{Z}^+):\pi=S_N+R_N$;
• the sequence $(S_N)_{N\in\mathbb{Z}_+}$ is strictly increasing and $\lim_{N\to\infty}S_N=\pi$. In particular, each $S_N$ is a better approximation of $\pi$ than the previous one.

Since$$(\forall n\in\mathbb N):\binom{2n}n<4^n=2^{2n},$$you have$$R_N<\sum_{n=N+1}^\infty\frac6{2^{2n+1}}=\frac1{4^N}.$$So, taking $N=0$, you get that $\pi=S_0+R_0$. But $S_0=3$ and $R_0<1$. So, the first digit of $\pi$ is $3$. If you take $N=3$, then $\pi=S_3+R_3$. But $S_3\approx3.14116$ and $R_3<0.015625$. So, the second digit is $1$. And so on…

The simplest method to explain to a child is probably the polygon method, which states that the circumference of a circle is bounded from below by the circumference of an inscribed regular $n$-polygon and from above by the circumference of a circumscribed polygon.

Once you have a bound from below and above, you can guarantee some digits. For example, any number between $0.12345$ and $0.12346$ will begin with $0.1234$.

I think the general answer you're looking for is:

Yes, proving that a method for calculating $\pi$ works requires also describing (and proving) a rule for when you can be sure of a digit you have produced. If the method is based on "sum such-and-such series", this means that one needs to provide an error bound for the series. Before you have that, what you're looking at is not yet a "method for calculating $\pi$".

So the answer to your first question is "Yes; because otherwise they wouldn't count as techniques for calculating $\pi$ at all".

Sometimes the error bound can be left implicit because the reader is supposed to know some general theorems that leads to an obvious error bound. For example, the Leibniz series you're using is an absolutely decreasing alternating series, and therefore we can avail ourselves of a general theorem saying that the limit of such a series is always strictly between the last two partial sums. Thus, if you get two approximations in succession that start with the same $n$ digits, you can trust those digits.

(The Leibniz series is of course a pretty horrible way to calculate $\pi$ -- for example you'll need at least two million terms before you have any hope of the first six digits after the point stabilizing, and the number of terms needed increases exponentially when you want more digits).

In other cases where an error bound is not as easy to see, one may need to resort to ad-hoc cleverness to find and prove such a bound -- and then this cleverness is part of the method.