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Explore Stack InternalLet $$a_0=3, a_n=\lfloor \pi a_{n-1} \rfloor$$ You have $a_n \in \mathbb{N}$ and $\frac{a_{n+1}}{a_n} \to \pi$
EDIT : Proof :
$$\forall n \in \mathbb{N}, a_n \pi \leq \lfloor \pi a_n \rfloor < a_n \pi+1$$$$\forall n \in \mathbb{N}, a_n \pi-1 \leq \lfloor \pi a_n \rfloor < a_n \pi+1$$ $$ \pi \leq \frac{a_{n+1}}{a_n} < \pi +\frac{1}{a_n}$$$$ \pi -\frac{1}{a_n} \leq \frac{a_{n+1}}{a_n} < \pi +\frac{1}{a_n}$$ and since $a_n \to \infty$ then$ \frac{a_{n+1}}{a_n} \to \pi$
Let $$a_0=3, a_n=\lfloor \pi a_{n-1} \rfloor$$ You have $a_n \in \mathbb{N}$ and $\frac{a_{n+1}}{a_n} \to \pi$
EDIT : Proof :
$$\forall n \in \mathbb{N}, a_n \pi \leq \lfloor \pi a_n \rfloor < a_n \pi+1$$ $$ \pi \leq \frac{a_{n+1}}{a_n} < \pi +\frac{1}{a_n}$$ and since $a_n \to \infty$ then$ \frac{a_{n+1}}{a_n} \to \pi$
Let $$a_0=3, a_n=\lfloor \pi a_{n-1} \rfloor$$ You have $a_n \in \mathbb{N}$ and $\frac{a_{n+1}}{a_n} \to \pi$
EDIT : Proof :
$$\forall n \in \mathbb{N}, a_n \pi-1 \leq \lfloor \pi a_n \rfloor < a_n \pi+1$$ $$ \pi -\frac{1}{a_n} \leq \frac{a_{n+1}}{a_n} < \pi +\frac{1}{a_n}$$ and since $a_n \to \infty$ then$ \frac{a_{n+1}}{a_n} \to \pi$
Let $$a_0=3, a_n=\lfloor \pi a_{n-1} \rfloor$$ You have $a_n \in \mathbb{N}$ and $\frac{a_{n+1}}{a_n} \to \pi$
EDIT : Proof :
$$\forall n \in \mathbb{N}, a_n \pi \leq \lfloor \pi a_n \rfloor < a_n \pi+1$$ $$ \pi \leq \frac{a_{n+1}}{a_n} < \pi +\frac{1}{a_n}$$ and since $a_n \to \infty$ then$ \frac{a_{n+1}}{a_n} \to \pi$
Let $$a_0=3, a_n=\lfloor \pi a_{n-1} \rfloor$$ You have $a_n \in \mathbb{N}$ and $\frac{a_{n+1}}{a_n} \to \pi$
Let $$a_0=3, a_n=\lfloor \pi a_{n-1} \rfloor$$ You have $a_n \in \mathbb{N}$ and $\frac{a_{n+1}}{a_n} \to \pi$
EDIT : Proof :
$$\forall n \in \mathbb{N}, a_n \pi \leq \lfloor \pi a_n \rfloor < a_n \pi+1$$ $$ \pi \leq \frac{a_{n+1}}{a_n} < \pi +\frac{1}{a_n}$$ and since $a_n \to \infty$ then$ \frac{a_{n+1}}{a_n} \to \pi$
