Simplest way to get the lower bound $\pi > 3.14$

Inspired from this answer and my comment to it, I seek alternative ways to establish $\pi>3.14$. The goal is to achieve simpler/easy to understand approaches as well as to minimize the calculations involved. The method in my comment is based on Ramanujan's series $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots\tag{1}$$ This is quite hard to understand (at least in my opinion, see the blog posts to convince yourself) but achieves the goal of minimal calculations with evaluation of just the first term being necessary.

On the other end of spectrum is the reasonably easy to understand series $$\frac\pi4=1-\frac13+\frac15-\cdots\tag2$$ But this requires a large number of terms to get any reasonable accuracy. I would like a happy compromise between the two and approaches based on other ideas apart from series are also welcome.

A previous question of mine gives an approach to estimate the error in truncating the Leibniz series $(2)$ and it gives bounds for $\pi$ with very little amount of calculation. However it requires the use of continued fractions and proving the desired continued fraction does require some effort.

Another set of approximations to $\pi$ from below are obtained using Ramanujan's class invariant $g_n$ namely $$\pi\approx\frac{24}{\sqrt{n}}\log(2^{1/4}g_n)\tag{3}$$ and $n=10$ gives the approximation $\pi\approx 3.14122$ but this approach has a story similar to that of equation $(1)$.

Solution 1:

From the elementary inequality $$\frac{\sin x}x\le\frac{2+\cos x}3,$$ we get with $x=\pi/6$ easily $\pi\ge\frac{18}{4+\sqrt{3}}=3.1402\ldots$
Proof of the inequality (elementary, though not obvious): let $$f(x)=\frac{\sin x}{x(2+\cos x)}.$$ In order to prove $f(x)\le\lim_{x\to+0}f(x)$, we prove $f(x)\le f(x/2)$. That follows from $$f(x)=f(x/2)\,\frac{(2+\cos x/2)\cos x/2}{1+2\cos^2 x/2},$$ since with $c=\cos x/2$, we have $$\frac{(2+c)c}{1+2c^2}=1-\frac{(1-c)^2}{1+2c^2}\le1.$$

Solution 2:

If we consider the Beuker-like integral $$ 0<\int_{0}^{1}\frac{x^8(1-x)^8}{1+x^2}\,dx = 4\pi-\frac{188684}{15015} $$ we get, through partial fraction decomposition and few operations in $\mathbb{Q}$, $$ \pi > \frac{47171}{15015} > 3.14159.$$

Inspired by Professor Vector's brilliant approach, I am adding a further approach.
By the Shafer-Fink inequality we have $\arctan(x)>\frac{3x}{1+2\sqrt{1+x^2}}$ for any $x>0$, hence by evaluating both sides at $x=\frac{1}{\sqrt{3}}$ we get $\pi>\frac{18}{13}(4-\sqrt{3})=3.140237\ldots$ A refinement of the previous inequality is $$\forall x>0,\qquad \arctan(x)>\frac{6x}{1+\sqrt{1+x^2}+2\sqrt{2}\sqrt{1+x^2+\sqrt{1+x^2}}} $$ and the evaluation at $x=\frac{1}{\sqrt{3}}$ produces the sharper bound $$ \pi > \frac{36}{2+\sqrt{3}+4 \sqrt{2+\sqrt{3}}} > 3.1415.$$

Yet another approach. The inequality $\arctan(x)>\frac{5x(21+11x^2)}{105+90x^2+9x^4}$ for any $x\in(0,1)$ comes from the Gauss continued fraction / the Padé approximants for the arctangent function. By replacing $x$ with $\frac{x}{1+\sqrt{1+x^2}}$, then evaluating at $x=\frac{1}{\sqrt{3}}$, we get the nice and tight approximation: $$ \pi > \color{blue}{\frac{5}{601}\left(944-327\sqrt{3}\right)}>3.141592.$$

There is also a nice geometric argument leading to an acceleration of Vieta's formula. Let $PQ$ be a side of a regular $n$-agon inscribed in a unit circle centered at $O$. Let $M$ be the midpoint of the minor arc $PQ$. We may consider the unique parabola through $P,M,Q$ and approximate the area of the circle sector delimited by $P,O,Q$ through $[POQ]$ plus the area of a parabolic segment, $\frac{4}{3}[PMQ]:$

It follows that if $A_n$ is the area of the inscribed $2^n$-agon, we have $$ \pi \geq A_n+\frac{4}{3}\left(A_{n+1}-A_n\right) = \frac{4}{3}A_{n+1}-\frac{1}{3}A_n $$ with $$ A_n = 2^{n-1} \sin\frac{\pi}{2^{n-1}} $$ and $\{A_n\}_{n\geq 2}$ being computable through a simple recursion, $$ A_{n+1} = 2^n \sqrt{\frac{1-\sqrt{1-\frac{A_n^2}{4^{2n-2}}}}{2}}$$ where $$ \frac{4}{3}A_5-\frac{1}{3}A_4 = \frac{4}{3}\left[8\sqrt{2-\sqrt{2+\sqrt{2}}}-\sqrt{2-\sqrt{2}}\right]=\color{green}{3.141}44\ldots $$ By considering a $12$-agon and a $24$-agon we get the simpler $$ \pi > 4\sqrt{2}(\sqrt{3}-1)-1 = \color{green}{3.141}10\ldots $$

Solution 3:

Take Machin's formula: $$\pi=16\tan^{-1}\frac15-4\tan^{-1}\frac1{239}$$ Expand the arctangents into their Taylor series: $$\pi=16\left(\color{blue}{\frac15-\frac1{5^3×3}}+\frac1{5^5×5}-\dots\right)-4\left(\color{blue}{\frac1{239}}-\frac1{239^3×3}+\dots\right)$$ $$=\color{blue}{\frac{16}5-\frac{16}{375}-\frac4{239}}+\delta$$ $$=\color{blue}{3.140596\dots}+\delta$$ with $0<\delta<\frac{16}{5^5×5}+\frac4{239^3×3}$ since the two series are alternating with term magnitudes strictly decreasing. This proves $\pi>3.140596\dots>3.14$.

Solution 4:

Similarly to Jack d'Aurizio's answer (first section), we have the simpler integral

$$\frac{1}{2}\int_0^1 \ \frac{x^3(1-x)^6}{1+x^2} dx = \pi-\frac{1759}{560}=\pi-\left(3.14+\frac{3}{2800}\right)$$

A similar one evaluating to exactly $$\pi-3.14=\pi-\frac{157}{50}$$ can be obtained with the methods of Lucas (http://educ.jmu.edu/~lucassk/Papers/more%20on%20pi.pdf)

Since

$$3.14=\frac{157}{50}=\frac{22}{7}-\frac{1}{350},$$

the result can be obtained subtracting Dalzell's integral

$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx =\frac{22}{7}-\pi$$

from $$\frac{4}{5}\int_0^1 x^3(1-x)^4 dx = \frac{1}{350},$$

which gives

$$\frac{1}{5}\int_0^1 \frac{x^3(1-x)^4(4-5x+4x^2)}{1+x^2} dx = \pi -\frac{157}{50} > 0 $$

This is a direct proof for $\pi>\frac{157}{50}$ because the integrand is non-negative in $(0,1)$, very close to this proof that $\frac{22}{7}$ exceeds $\pi$.

The corresponding series is

$$1152\sum_{k=0}^\infty \frac{2k^2+8k+13}{(4k+4)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+12)}=\pi-\frac{157}{50}$$

(check

We can also take the first term out of a series relating $\pi$ with the approximation from below $\frac{25}{8}$ to get

$$\begin{align} \pi &= \frac{25}{8} + \sum_{k=0}^\infty \frac{360}{(4k+2)(4k+4)(4k+5)(4k+7)(4k+8)(4k+10)}\\ &=\frac{1759}{560} + \sum_{k=1}^\infty \frac{360}{(4k+2)(4k+4)(4k+5)(4k+7)(4k+8)(4k+10)}\\ \end{align}$$

where $\frac{1759}{560}>3.14$ as shown above.

This fraction already appeared in the first integral, but proving the series as in this answer now gives

$$\frac{1}{4}\int_0^1 \frac{x^5(1-x)^4(1+4x+x^2)}{1+x^2}dx = \pi -\frac{1759}{560}$$

Solution 5:

For modest requirements of accuracy, we can use a quite basic and easy to understand convergence acceleration technique to transform the Gregory/Leibniz series into something that yields the result with little computation.

If we have an alternating series

$$\sum_{n = 0}^{\infty} (-1)^n a_n$$

where the $a_n$ are slowly converging to $0$, then it takes little guessing to believe that the arithmetic mean of two successive partial sums of the series is a much better approximation to the value of the series than either or the two partial sums. We can write the mean as

$$\sum_{n = 0}^{m-1} (-1)^n a_n + \frac{(-1)^m a_m}{2} = \frac{a_0}{2} + \frac{1}{2} \sum_{n = 0}^{m-1} (-1)^n(a_n - a_{n+1})$$

and see that if $\bigl(a_n - a_{n+1}\bigr)_{n\in \mathbb{N}}$ is a decreasing sequence we immediately have an error bound of $\frac{1}{2}(a_{m} - a_{m+1})$ for this arithmetic mean. If $(a_n)$ is slowly converging, then this is much smaller than either of $a_m$ and $a_{m+1}$. For nice $(a_n)$, this can be iterated.

If we apply that to $a_n = \frac{1}{2n+1}$, we find $a_n - a_{n+1} = \frac{2}{(2n+1)(2n+3)}$, which is decreasing - and still converges slowly to $0$, so we get a significant improvement from then applying the method again. We get

\begin{align} \frac{\pi}{4} &= \sum_{n = 0}^{\infty} \frac{(-1)^n}{2n+1} \\ &= \frac{1}{2} + \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)} \\ &= \frac{1}{2} + \frac{1}{6} + 2\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)} \\ &= \frac{1}{2} + \frac{1}{6} + \frac{1}{15} + 6\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)(2n+7)} \end{align}

and so on, each series converging faster than the previous. Since the terms get more complicated, it is computationally - at least when done by hand - better to start accelerating not right at the beginning of the series, but rather compute a partial sum of the original series, and then use acceleration only on the remainder. Let's say that starting the remainder at $n = 4$ wouldn't need an unreasonable amount of computation, then we get

\begin{align} \frac{\pi}{4} &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \sum_{n = 4}^{\infty} \frac{(-1)^n}{2n+1} \\ &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)} \\ &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + 2\sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)} \\ &= 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + \frac{1}{1287} + 6\sum_{n = 4}^{\infty} \frac{(-1)^n}{(2n+1)(2n+3)(2n+5)(2n+7)} \\ &> 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{18} + \frac{1}{198} + \frac{1}{1287} \\ &> 0.78519 \end{align}

and thus $\pi > 3.14076$.

Of course Machin-like formulæ are far more efficient. But it's not too terrible.