How can $\pi$ be irrational if it can be calculated using an infinite series of rational numbers?

Question

Just a curiosity. I just saw a video where pi was being calculated by the series

$$1-(1/3)+(1/5)-(1/7)+...$$

My question is how can $\pi$ be irrational if we can calculate it using rational numbers?

Answer 1

A sequence of rational numbers can converge to an irrational. Yes, all the partial sums are rational but the limit need not be. There is nothing special about $\pi$ here, it is true of all irrationals. In fact, the most common construction of the real numbers from the rationals is through Dedekind cuts, which separate the rationals into the sets of those above and below the irrational. You can then find a sequence of rationals in either set that converges to the irrational.

Answer 2

For a simple example: any real number $x\,$, rational or not, has a decimal representation. Consider now the sequence $x_n$ defined by truncating the representation at the $n^{th}$ decimal digit, then each $x_n$ is obviously a rational (since it has a finite decimal representation) and $x_n \to x\,$.

In the case of $\pi = 3.1415\dots\,$, the sequence would start as $3, \,3.1, \,3.14, \,3.141, \,3.1415 \dots \to \pi$.