Find an increasing sequence of rationals that converges to $\pi$ I am not sure how to construct a sequence that would convey convergence to $\pi$.
Except maybe $a_n=\{\pi + 1/n\}$ but the terms would not be rational.
Looking for an adequate way to show to satisfy the three conditions.

Leibniz formula tells us:
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1}=\frac{\pi}4=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\dots$$
Group the terms by two, the difference is positive, so there is an obvious increasing sequence of rationals that converges to $\pi$:
$$4\left(1-\frac{1}{3}\right)+4\left(\frac{1}{5}-\frac{1}{7}\right)+\dots$$
Or more formally
$$S_n=4\sum_{k=0}^{n} \left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)=8\sum_{k=0}^{n}\frac{1}{(4k+1)(4k+3)}$$
Then $S_n$ is obviously rational and increasing, and $S_n\to\pi$.
