Exercise 8.3.4

Question

Show
$$\lim_{n \to \infty} \frac{b_{2n}}{b_{2n+1}} = 1,$$
and use this fact to finish Wallis's product formula in (3).

Ulisse Mini · Accepted Answer

Equivalently, we wish to evaluate
$$\lim_{n \to \infty} \frac{b_n}{b_{n+1}} = \lim_{n \to \infty} \frac{\int^{\pi/2}_0 \sin^{n+1} (t) dt}{\int^{\pi/2}_0 \sin^n (t) dt}$$
Note that since $b_n > b_{n+1}$, $\frac{b_n}{b_{n+1}} \geq 1$.
We prove the following lemma: for $\frac{\pi}{2} > b > a > 0$,
$$\lim_{n \to \infty} \frac{\int^{\pi/2}_b \sin^n x}{\int^a_0 \sin^n x} = \infty$$

To see this, let $y_n$ be the value in the limit. Note that $\int^{\pi/2}_b \sin^{n+1}x \geq \sin b \int^{\pi/2}_b \sin^n x$ and similarly $\int_0^a \sin^{n+1}x \leq \sin a \int_0^a \sin^n x$, and note
$$\frac{y_{n+1}}{y_n} = \left(\frac{\int^{\pi/2}_b \sin^{n+1} x}{\int^{\pi/2}_b \sin^{n} x}\right) \left(\frac{\int_0^a \sin^{n} x}{\int_0^a \sin^{n+1} x}\right) \geq \frac{\sin b}{\sin a}> 1$$
In other words, $(y_n)$ grows at least exponentially, and therefore must not be bounded.

We now show that for arbitrary $\epsilon > 0$, we can find $N$ large enough so that for $n \geq N$,
$$\int^\frac{\pi}{2}_0 \sin^{n+1} x \geq (1-\epsilon) \int_0^\frac{\pi}{2} \sin^n x$$
Note that, for any $a \in (0, \pi / 2)$,
$$\int^\frac{\pi}{2}_0 \sin^{n+1} x = \int_0^a \sin^{n+1} x + \int_a^\frac{\pi}{2} \sin^{n+1} x \geq \sin a \int_a^\frac{\pi}{2} \sin^n x$$
and for any $b \in (a, \pi/2)$,
$$\int_0^\frac{\pi}{2} \sin^n x = \int_0^a \sin^n x + \int_a^b \sin^n x + \int_b^\frac{\pi}{2}\sin^n x$$
Find $a \in (0, \pi/2)$ so that $\sin a > \sqrt{1-\epsilon}$, and fix $b = \left(a + \frac{\pi}{2}\right)/2$. From our earlier lemma, we can find $N$ large enough that $n > N$ ensures
$$\int_0^a \sin^n x < \epsilon_1 \int_b^\frac{\pi}{2}\sin^n x$$
where we'll choose $\epsilon_1$ to satisfy $\frac{1}{1 + \epsilon_1} = \sqrt{1 - \epsilon}$. This ensures that
$$\int^\frac{\pi}{2}_a \sin^n x > \sqrt{1-\epsilon} \int_0^\frac{\pi}{2} \sin^n x$$
and
$$\int^\frac{\pi}{2}_0 \sin^{n+1} x \geq (1-\epsilon) \int_0^\frac{\pi}{2} \sin^n x.$$
We can convert this to indicate that for large enough $n$, $b_n / b_{n+1} \leq 1 + \epsilon$, which together with our earlier observation that $b_n > b_{n+1}$ lets us conclude
$$\lim_{n \to \infty} \frac{b_n}{b_{n+1}} = 1$$
On the other hand, plugging in the formulas for $b_{2n}$ and $b_{2n+1}$ in Exercise 8.3.3 (c) into
$$\lim_{n \to \infty} \frac{b_{2n}}{b_{2n+1}}$$
leaves us with Wallis's product.

Exercise 8.3.4

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Exercise 8.3.4

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