A formula for $\pi$ and an inequality

Question

For any $n\in \mathbb{N}$ prove the identity : $$\pi =\sum_{k=1}^{n}\frac{2^{k+1}}{k\dbinom{2k}{k}}+\frac{4^{n+1}}{\dbinom{2n}{n}}\int_{1}^{\infty}\frac{\mathrm{d}x}{(1+x^2)^{n+1}}\tag{1}$$ and thus derive the inequality : $$\frac{2}{2^n\sqrt{n}}<\pi -\sum_{k=1}^{n}\frac{2^{k+1}}{k\dbinom{2k}{k}}<\frac{4}{2^n\sqrt{n}}\tag{2}$$ How to prove these two? I was given them in my calculus exam. I wasted 2 hours of the 4 hour allotted on them but nothing came to my mind. Can someone help? Thanks a lot.

Answer 1

Let

$$I_n = \int_1^\infty \frac{dx}{(1+x^2)^{n+1}}.$$

Then $I_0 = \frac{\pi}{4}$ is well-known, and $(1)$ holds for $n = 0$. Integration by parts yields

$$\begin{align} I_n &= \int_1^\infty \frac{dx}{(1+x^2)^{n+1}}\\ &= \left[\frac{x}{(1+x^2)^{n+1}}\right]_1^\infty + 2(n+1) \int_1^\infty \frac{x^2\,dx}{(1+x^2)^{n+2}}\\ &= -\frac{1}{2^{n+1}} + 2(n+1) \left[I_n - I_{n+1}\right]. \end{align}$$

Rearranging yields

$$I_n = \frac{2(n+1)}{2n+1}I_{n+1} + \frac{1}{(2n+1)2^{n+1}},$$

and so

$$\begin{align} \frac{4^{n+1}}{\binom{2n}{n}} I_n &=\frac{4^{n+1}}{2^{n+1}(2n+1)\binom{2n}{n}} + \frac{4^{n+1}}{\binom{2n}{n}\frac{2n+1}{2(n+1)}}I_{n+1}\\ &= \frac{2^{n+2}}{\frac{(n+1)2(n+1)(2n+1)}{(n+1)(n+1)}\binom{2n}{n}} + \frac{4^{n+2}}{\binom{2n}{n}\frac{(2n+1)2(n+1)}{(n+1)(n+1)}}I_{n+1}\\ &= \frac{2^{n+2}}{(n+1)\binom{2(n+1)}{n+1}} + \frac{4^{n+2}}{\binom{2(n+1)}{n+1}}I_{n+1}, \end{align}$$

which completes the induction on $(1)$.

Then $(2)$ is equivalent to

$$\frac{2}{2^n\sqrt{n}} < \frac{4^{n+1}}{\binom{2n}{n}}I_n < \frac{4}{2^n\sqrt{n}}.$$

Here, we must assume $n > 0$ of course.

We can estimate

$$I_n < \int_1^\infty \frac{x\,dx}{(1+x^2)^{n+1}} = \left[-\frac{1}{2n(1+x^2)^n}\right]_1^\infty = \frac{1}{n2^{n+1}}$$

and

$$I_n > \int_1^\infty \frac{2x\,dx}{(1+x^2)^{n+2}} = \left[-\frac{1}{(n+1)(1+x^2)^{n+1}}\right] = \frac{1}{(n+1)2^{n+1}}.$$

The former estimate shows that it is enough to prove

$$\frac{2^{2n-1}}{\sqrt{n}} \leqslant \binom{2n}{n}\tag{$\ast$}$$

to obtain the upper bound. For $n = 1$, we have equality in $(\ast)$, and then

$$\binom{2n+2}{n+1} = \binom{2n}{n}\frac{4(n+\frac{1}{2})}{n+1} \geqslant \frac{2^{2n-1}}{\sqrt{n}} \frac{4(n+\frac{1}{2})}{n+1} = \frac{2^{2n+1}}{\sqrt{n+1}}\frac{n+\frac{1}{2}}{\sqrt{n(n+1)}} > \frac{2^{2n+1}}{\sqrt{n+1}}$$

completes the induction for $(\ast)$.

For the lower bound, it suffices to show

$$\binom{2n}{n} \leqslant \frac{\sqrt{n}2^{2n}}{n+1}\tag{$\ast\ast$}.$$

For $n = 1$, we have equality in $(\ast\ast)$, and then

$$\begin{align} \binom{2(n+1)}{n+1} & \leqslant \frac{\sqrt{n}2^{2n}}{n+1}\frac{(2n+1)2(n+1)}{(n+1)^2}\\ &= \frac{2^{2n+2}\sqrt{n}(n+\frac{1}{2})}{(n+1)^2}\\ &= \frac{2^{2n+2}\sqrt{n+1}}{n+2}\frac{\sqrt{n}(n+2)(n+\frac{1}{2})}{(n+1)^{5/2}}\\ &< \frac{2^{2n+2}\sqrt{n+1}}{n+2} \end{align}$$

finishes the proof.