Is $\pi$ approximately algebraic?

Question

As we know, $\pi$ is transcendental, meaning that there is no rational numbers $a_0,\ldots,a_n\in\mathbb{Q}$ such that $$a_0+a_1\pi+\cdots+a_n\pi^n=0.$$ But I was wondering if we can get this as a limiting process: Is there a sequence of polynomials $\{p_n(x)\}_{n=1}^\infty$ with rational coefficients such that the first positive root of $p_n(x)$ tends to $\pi$ as $n\to\infty$?

Answer 1

Yes, $$\sin\pi=0$$ so the first positive root of $$p_n(x)=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots+\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ tends to $\pi$ as $n\to\infty$.