Doubt in Ivan Niven's proof of irrationality of pi.

Question

In the proof, how do we get the upper limit for $f(x) \sin{x}$ as $\pi^n \cdot \frac{a^n}{n!}$ ?

I thought $f(x) \sin{x}$ would be maximum at $x=\pi/2$ when its value would be: $$\pi^n \cdot \frac{a^n}{2^{2n}n!}$$

Where am I going wrong?

Answer 1

The upper limit of $f(x) \sin{x}$ mentioned in the proof is not the least upper bound.

The least upper bound is $$\pi^n \cdot \frac{a^n}{2^{2n}n!}$$ as mentioned in the question.

Given $f(x) = \frac{x^n.(a-bx)^n}{n!}$ a general argument for getting the upper bound mentioned in the proof would be that the maximum value of $x$ is $\pi$ and the maximum value of $(a-bx)$ is $a$.

After posting this question here, I found an explanation of the limits of the proof in Presh Talwalkar's blog

I posted my doubts in the comments section and Presh was kind enough to respond and clarify my doubt.