Concluding a proof ($\pi$ is irrational) I am making a proof for irrationality of $\pi$ and i proceeded as follows:
Let $\pi=\frac{u}{v}$ for some $u,v\in \Bbb{N}$, define family of integrals:
$$I_n=\frac{v^{2n}}{n!}\int_0^\pi x^n(\pi-x)^n \sin x\,dx$$ By some elementary estimates we have
$$0<I_n\leq\frac{v^{2n}}{n!}\pi^{2n+1}$$ thus by squeeze lemma we have $\lim_{n\to\infty}I_n=0$.
In the next part I prove that $I_0,I_1\in \Bbb{N}$ and applying some integration by parts i get recursive formula $$I_n=(4n-2)v^2I_{n-1}-u^2v^2I_{n-2}\tag{1}$$
Now I should conclude the proof and I have two ideas but I'm not really sure whether both of them are correct (if both, which one is better?):
a) Because of (1), we can say that $\lim_{n\to\infty}I_n=\infty$ because of the factor $4n-2$ thus implying there are two different limits for $I_n$ which is impossible thus contradiction.
b) Because $I_0,I_1\in \Bbb{N}$. Then because $u,v\in\Bbb{N}$ too, recursive formula shows that $\forall n\in\Bbb{N}:I_n\in\Bbb{N}$ but one cannot have infinite sequence of natural numbers tending to zero from above (obviously they can't approach from below).

Good question! Given the fact that your estimate and (1) are correct, you should choose (b) rather than (a) because the latter is not correct.
As @InterstellarProbe has mentioned in the comment, there are some sequences $a_n$ tending to $0$, but with $(4n-2)a_n$ still tending to zero$.
The point is how fast $a_n$ tends to zero. If it tends to zero "faster than" $(4n-2)$ tends to $\infty$, then their product $(4n-2)a_n$ still tends to zero.
