Someone in discussion asked, and we answered…
We know a formula for $\pi$ as
\[f(\pi) = \sum (-1)^{x} \frac{1}{2x + 1}\]but suppose we loosen the criterion as follows:
\[g(\pi) = \sum (-1)^x \frac{1}{3x + 1}\]how can we prove this too be a function evaluated at $\pi$? The first one is $f(x) = x/4$ evaluated at $\pi$. How about the second? It’s okay to be wrong you know.
You can also see the discussion notes right here!
(example of easy links reference).
