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.
 A: To build on my comment, there are probably a few ways to define that sequence. One "lazy" way to do it is to recognize that $\pi$ as an irrational has the property that for any $n \in \mathbb{N}$, we can find the $n^{th}$ decimal in $\pi$'s decimal expansion. So for $b_i \in \{0,1,\ldots,8,9\}$ we can write $$\pi = b_0.b_1b_2b_3b_4\ldots b_{k-1}b_kb_{k+1}\ldots$$ where obviously $b_0 = 3, b_1 = 1$, etc. Now define $a_n = b_0.b_1\ldots b_n$ to get the sequence $$\{3,3.1,3.14,\ldots \}$$ Along with this you may also need to prove that $a_k \leq a_{k+1}$ for all $k\in \mathbb{N}$ but that should be pretty easy if you just consider $a_{k+1}-a_{k}$.
A: The sequence in graydad's answer can be written in terms of the floor function as
$$n \mapsto 10^{-n} \lfloor 10^n \pi \rfloor.$$
This suggests some cute generalizations. Instead of the decimal expansion for $\pi$, we can use the expansion in any base $q$:
$$n \mapsto q^{-n} \lfloor q^n \pi \rfloor$$
converges to $\pi$ for any whole number $q \ge 2$. In fact,
$$n \mapsto a_n^{-1} \lfloor a_n \pi \rfloor$$
converges to $\pi$ for any increasing sequence $n \mapsto a_n$ of natural numbers!
A: Let $n_0:=\lfloor\pi\rfloor$ and define inductively $$n_{k+1}:=\left\lfloor10^{k+1}\left(\pi-\sum_{i=0}^k\dfrac{n_i}{10^i}\right)\right\rfloor.$$ Then $$\lim_{m\to\infty}\sum_{k=0}^m\dfrac{n_k}{10^k}=\sum_{k=0}^\infty\dfrac{n_k}{10^k}=\pi.$$
A: 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 $S_n\to\pi$.
