Exercise 8.3.5

Question

Derive the following alternative form of Wallis's product formula:
$$\sqrt{\pi} = \lim_{n \to \infty} \frac{2^{2n} (n!)^2}{(2n)! \sqrt{n}}$$

Ulisse Mini · Accepted Answer

$$ \begin{aligned}
    \frac{\pi}{2} &= \prod^\infty_{i=1} \frac{(2i)^2}{(2i-1)(2i+1)} = \lim_{n \to \infty} \frac{2^n (n!)^2}{\left(\prod^n_{i=1} (2i-1)\right)\left(\prod^n_{i=1} (2i + 1) \right)} \
    &= \lim_{n \to \infty} \frac{2^{2n} (n!)^2 (2^n n!) \left(2^n n! (2n + 1)\right)}{\left((2n+1)!\right)^2} = \lim_{n \to \infty} \left(\frac{2^{2n} (n!)^2}{(2n)!}\right)^2 \left(\frac{1}{2n}\right)\left(\frac{2n}{2n+1}\right) \
    \pi &= \left(\lim_{n \to \infty} \left(\frac{2^{2n} (n!)^2}{(2n)!}\right)^2 \left(\frac{1}{n}\right) \right) \left( \lim_{n \to \infty}\frac{2n}{2n+1}\right) =\left(\lim_{n \to \infty} \frac{2^{2n} (n!)^2}{(2n)! \sqrt{n}}\right)^2  \
\sqrt{\pi} &= \lim_{n \to \infty} \frac{2^{2n} (n!)^2}{(2n)! \sqrt{n}}
\end{aligned}
   $$

Exercise 8.3.5

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Exercise 8.3.5

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