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 $(1)$.

  • 2
    $\begingroup$ Probably not what you are looking for. Machin's formula becomes accurate enough with a modest number of terms. Even the series of $\arctan(1/5)$ converges fast enough. Just trying to narrow down the target :-) $\endgroup$ Oct 23, 2017 at 6:55
  • 4
    $\begingroup$ Why the downvote? If this appears to be a low quality question, please suggest ways to improve. I have not asked too many questions but I do try my best to improve the quality of my questions. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 7:09
  • 2
    $\begingroup$ Another downvote? Perhaps today is my lucky day to have a good supply of wonderful answers and downvotes too. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 8:17
  • 2
    $\begingroup$ Have a look at math.ucla.edu/~vsv/resource/general/… $\endgroup$ Oct 23, 2017 at 8:33
  • 1
    $\begingroup$ If there is an easy way to show, say, $\log(\pi)>\frac{103}{90}$, exponentiating gives $e^{\log(\pi)}=\pi>e^\frac{103}{90}$, which proves your inequality with seven terms of the McLaurin series. $\endgroup$ Oct 26, 2017 at 9:03

14 Answers 14


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 $$\frac{(2+c)c}{1+2c^2}=1-\frac{(1-c)^2}{1+2c^2}\le1.$$

  • 13
    $\begingroup$ How did you arrive at (think of) this inequality? I have never seen it even as a part some typical exam/exercise. +1 and this one appears very very simple. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 7:43
  • 7
    $\begingroup$ Your method of proof is so novel and elegant that I had to comment again. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 8:34
  • 6
    $\begingroup$ @Paramanand Singh It's certainly not a novel method. I don't remember where I've seen it, but it was more than 20 years ago. One can use it to prove $$\frac{\sin x}x\ge\sqrt[3]{\cos x}$$ for $|x|\le\pi/2$, too. With $x=\pi/6$, this gives $\pi\le\sqrt[6]{972}=3.1473\ldots$ $\endgroup$
    – user436658
    Oct 23, 2017 at 9:44
  • 3
    $\begingroup$ (+1) Nice solution, but Huygens' inequality is just an instance of the Shafer-Fink inequality! (thanks for the idea, by the way). $\endgroup$ Oct 23, 2017 at 11:31

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, $\frac{4}{3}[PMQ]:$

enter image description here

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 $$ \pi > 4\sqrt{2}(\sqrt{3}-1)-1 = \color{green}{3.141}10\ldots $$

  • 9
    $\begingroup$ How many integrals are stored in your memory?? :) I believe the integral gets evaluated easily by expressing integrand as a polynomial + a rational function. +1 $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 7:32
  • 1
    $\begingroup$ The approximation $(18/13)(4-\sqrt{3})$ is good enough and matches the one given by Professor Vector but you get it through a different inequality. The more amazing aspect is that you have a refinement also!! Your last inequality for $x=1$ gives the desired conclusion $\pi>3.14$ $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 11:36
  • 3
    $\begingroup$ Don't know what to say now. Looks like you have $n$ number of weapons in your arsenal where $n\to\infty$ with growing time. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 13:23
  • 3
    $\begingroup$ Coming home from work and observing that my favorite moderator has already presumptuously used both approaches i could imagine to answer this wonderful question is really frustrating. (+1) for sure! $\endgroup$
    – tired
    Oct 23, 2017 at 18:25
  • 1
    $\begingroup$ A small change to your first integral gets the precision of your last approach: $$\frac{\sqrt{3}}{27}\int_0^1 \frac{x^8(1-x)^8}{1+x+x^2}dx = \pi - \frac{17647759\sqrt{3}}{9729720}$$ $\endgroup$ Oct 25, 2017 at 22:36

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 $\pi>3.140596\dots>3.14$.

  • $\begingroup$ This meets both the criteria : formula is simple to understand and calculations are also easy (doable by hand within minutes). +1 Will wait for more answers before accepting any. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 7:28

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)



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

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


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

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


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.

  • $\begingroup$ This is a good approach. +1 While it does require some effort but still doable by hand in few minutes. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 11:09
  • 1
    $\begingroup$ I like the use of acceleration. $\endgroup$ Oct 23, 2017 at 19:13
  • $\begingroup$ Typo alert! I think sign of $\frac13$ should be negative. $\endgroup$
    – karakfa
    Mar 5, 2018 at 2:54
  • $\begingroup$ @karakfa Indeed. Thanks for spotting it. $\endgroup$ Mar 5, 2018 at 12:09
  • $\begingroup$ This is a special case of Euler Summation. $\endgroup$
    – Kapil
    Jul 30, 2020 at 4:27

Another simple method is to use the Newton-Raphson method to find the zero of the function $\sin(x)$ at $x = \pi$ using the initial guess of $x = 3$. The fact that at $x = \pi$ the function $\sin(x)$ has a point of inflexion, means that convergence to $x = \pi$ will happen in an alternating way, therefore you obtain rigorous upper and lower bounds. So, the sequence defined by the recurrence:

$$x_{n+1} = x_n - \tan(x_n)$$

and $x_0 = 3$ converges to $\pi$. We have $x_1 = 3.1425\cdots$, $x_2 = 3.1415926533\cdots$, and therefore $\pi > 3.1415926533$. Now, we need to compute $x_2$ here without using $\pi$ itself, we can use the series expansions of $\sin(x)$ and $\cos(x)$ around $x = 0$. These are both alternating series, so we can easily get to an accurate bound on $x_1$ and from that we can get to an accurate bound on $x_2$, the lower bound of which then yields an accurate lower bound for $\pi$.

  • 3
    $\begingroup$ If the goal is just to compare $3.14$ to $\pi$, it's even faster to start the iteration at $x_0=3.14$ and simply see which way it moves. Then we don't even need to compute the cosine in the denominator, because we know already what its sign will be. $\endgroup$ Oct 23, 2017 at 14:40
  • 1
    $\begingroup$ In fact, the quadratic convergence of Newton-Raphson ought to hold even if we replace $f'(x_n)$ in the recurrence with $f'(x_{\rm root})$ (which in this case we happen to know exactly), so we can iterate $x\mapsto x + \sin x$ instead and save a series evaluation and a long division in each step. The convergence will then not be alternating, but we can still get a bound from the opposite side as $x_{n-1}+2\sin x_{n-1}$, since as soon as we're even slightly close we know $f'(x_n)\in [-1,-\frac12)$. $\endgroup$ Oct 23, 2017 at 14:55

Since $\cos \frac\pi3 = \frac12$ and the cosine is strictly decreasing between $0$ and $\pi$, we can prove that $3.14 < \pi$ by showing that $\cos\frac{3.14}3 > \frac12$. This takes only 4 terms of the power series for cosine:

x^0/0! = 1   adding gives 1
x^2/2! = 0.547755555555556   subtracting gives 0.452244444444444
x^4/4! = 0.0500060247736625   adding gives 0.502250469218107
x^6/6! = 0.00182607185873483   subtracting gives 0.500424397359372

Since the series is alternating and absolutely decreasing, when we have just subtracted a term and gotten something above $\frac12$ we now know that the limit is larger than $\frac12$.

Edit: showing that $\sin\frac{3.14}6 < \frac12$ is slightly faster:

x^1/1! = 0.523333333333333   adding gives 0.523333333333333
x^3/3! = 0.0238882283950617   subtracting gives 0.499445104938272
x^5/5! = 0.000327122745394376   adding gives 0.499772227683666

One can do this with pen and paper, keeping 4 digits after the decimal point in every intermediate (and explicit interval arithmetic), using just two 4-by-4 multiplications and two 4-by-3 to show $3.141<\pi$:

handwritten calculation

Five terms to 10 decimal digits will show that $3.14159 < \pi < 3.14160$.

  • 2
    $\begingroup$ The arctan series has become so popular that one misses the usual series for sine / cosine. Very direct and simple approach +1 $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 13:18
  • $\begingroup$ OMG!! Indisputable proof that it is calculated via hand :) :) however I was convinced by your earlier version too. $\endgroup$
    – Paramanand Singh
    Oct 24, 2017 at 3:28
  • $\begingroup$ In my high school days when there was no computer / internet and calculator could give only 10 digits, I used lot of paper to get hold of digits of $e$(simple) and $\pi$(not so simple). Got reminded of those days with your answer. $\endgroup$
    – Paramanand Singh
    Oct 24, 2017 at 3:33

A very easily understandable way is to use regular polygons inscribed into a circle of radius $1$. If $a_n$ is the length of the side of the $n$-gon, then $$a^2_{2n} = 2-\sqrt{4-a^2_n}\text{.}$$  A lower bound on $a_n$ implies a lower bound on $a_{2n}$, we can thus round down intermediary results. Four applications of the formula with 4 to 5 significant digits are enought to establish the bound of $3.14$. $$\begin{array}{rl} a^2_6 =& 1\\ a^2_{12} >& 0.2679 \\ a^2_{24} >& 0.06813 \\ a^2_{48} >& 0.017105 \\ a^2_{96} >& 0.004280 \\ \end{array}$$ Leading to $a_{96} > 0.06542$ and $\pi > 48a_{96} > 3.14$.

  • $\begingroup$ This is the approach given by David Mitra in comments to my question. His linked answer contains more details. But since you added it here on your own +1 for you. $\endgroup$
    – Paramanand Singh
    Oct 23, 2017 at 11:55
  • 3
    $\begingroup$ This is basically how Archimedes showed $\pi > 3+\frac{10}{71}$. $\endgroup$
    – Sil
    Oct 23, 2017 at 11:56

This is a slight variant on Henning Makholm's answer, with an emphasis on actually carrying out the computations that are required to verify the desired inequality.

Let $u=.314$. It suffices to show that $\sin u\lt\sin\left({\pi\over10}\right)={\sqrt5-1\over4}$. But

$$\sin u\lt u-{u^3\over6}+{u^5\over120}\lt u\left(1-{u^2\over6}+\left({u^2\over6}\right)^2\right)$$

At this point it's not too hard to get a calculator to tell you that


but let's do it showing (almost) all the arithmetic to establish the inequalities


The second inequality is equivalent to $2.236\lt\sqrt5$. Since $2236/4=559$, this is equivalent to $559^2\lt5\cdot250^2=312500$. It's not too hard (especially if you use a calculator...) to compute $559^2=312481$, which verifies the inequality. (Note, I consider $5\cdot25^2=5\cdot625=3125$ to be an "easy" multiplication.)

As for the first inequality, since $.309=.314-.005$, it is equivalent to


The (not too hard) computation $314^2=98596$ followed by division by $6$ shows that


It follows that


where the final inequality is verified by $2\cdot314=628\gt625=5\cdot125$.

If there is an easier way to organize the arithmetic so that you don't have to do a couple of messy multiplications (e.g., $559\times559$ and $314\times314$) somewhere along the line, I'd like to see it. (Even Professor Vector's marvelous answer, when ${18\over4+\sqrt3}\gt3.14={157\over50}$ is simplified to the inequality $900-628=272\gt157\sqrt3$, seems to require computing $272^2$ and $157^2$.)

  • 1
    $\begingroup$ Since 3.14 already has 3 significant digits, I think it is not unreasonable to need some arithmetic on numbers of that size. Computing 3 terms of $\sin (3.141/6)$ with a precision of $10^{-4}$ in every intermediary requires two 4-by-4 multiplications and two 4-by-3, hardly onerous even with pencil and paper. $\endgroup$ Oct 23, 2017 at 21:45

From $$\pi=\sum_{k=0}^{+ \infty}{\dfrac {1}{16^k}} \cdot \dfrac {120k^2+151k+47}{512k^4+1024k^3+712k^2+194k+15}$$ you obtain your inequality by summing just the first two terms.

  • $\begingroup$ The formula used here is the famous Bailey-Borwein-Plouffe formula which can be proved using integrals. $\endgroup$
    – Paramanand Singh
    Oct 24, 2017 at 3:57
  • $\begingroup$ @ParamanandSingh Yes, maybe this is not in the category of simplest approaches, but, as you see, if we discard addition in the calculation of digits, then there are only two divisions to perform. $\endgroup$
    – user480281
    Oct 24, 2017 at 12:16
  • $\begingroup$ And all the terms are strictly positive so that guarantees us that only the first two steps are needed. $\endgroup$
    – user480281
    Oct 24, 2017 at 12:19
  • $\begingroup$ I am aware of the proof of BBP and the proof is not difficult (integral evaluation of a rational function) so I am counting it as simple. And +1 has been delivered to you already. :) $\endgroup$
    – Paramanand Singh
    Oct 24, 2017 at 17:14
  • $\begingroup$ @ParamanandSingh Haha thanks, I would also like to mention that this approach with BBP is probably more simple then yours with Ramanujan´s formula because if that formula alternates with + and - signs there is also a need to show that only the first term suffices, that is, that the negative terms will not take over some decimals. $\endgroup$
    – user480281
    Oct 24, 2017 at 17:17

Equation (2) in the question is the result of plugging $x=1$ in the expansion for the arctangent, using


Instead, from $$\frac{\pi}{6}=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)$$ a series with faster convergence is obtained. Taking six terms,


Four terms from $$\frac{\pi}{8}=\tan^{-1}\left(\sqrt{2}-1\right),$$ give

$$\pi> 8\left(\sqrt{2}-1-\frac{(\sqrt{2}-1)^3}{3}+\frac{(\sqrt{2}-1)^5}{5}-\frac{(\sqrt{2}-1)^7}{7}\right)=\frac{32}{105}\left(716-499\sqrt{2}\right)>3.141,$$

while two terms from $$ \frac{\pi}{12}=\sin^{-1}\left( \frac{\sqrt{3}-1}{2\sqrt{2}} \right)$$

lead to $$\pi > 12\left( \frac{\sqrt{3}-1}{2\sqrt{2}} + \frac{1}{6} \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^3 \right) = \frac{27\sqrt{3}-29}{4\sqrt{2}}>3.14$$


From series $$\pi^6=960\sum_{k=0}^\infty \frac{1}{(2k+1)^6},$$


But $3.14^6<960$, so $3.14^6<\pi^6$ and therefore $3.14<\pi$.


Do you want a solution that is a compromise between calculation efficiency and ease of understanding? This story might help.

Many, many years ago in school, I was introduced to Fortran and had limited access to a computer. It was physically large but very low power by today's standards. I knew the famous $\frac{\pi}{4} = \tan^{-1}(1)$ formula but I also realised how slowly it converges. I would not be able to get far with the run time available to me.

The internet did not exist yet and the school's and the local public library did not help. I knew how the $\frac{\pi}{4}$ was derived so I played with other trig formulae. I managed to calculate the Taylor series of $\sin^{-1}$ by a mixture of messy differentiation and induction. I figured that if I evaluated $\frac{\pi}{6} = \sin^{-1}(\frac{1}{2})$, the convergence would be linear: twice the terms would give me twice the number of decimal places. In practice, the algorithm was quadratic since if I aimed at $n$ times as many decimal places, I would need $n$ times as many terms and each calculation would take $n$ as long.

I forget the run time that I had but the best I achieved was 500 decimal places.

Many years later but still long ago, I rewrote the program in C and ran it on an idle Unix system at work. In a month, it calculated a million decimal places. For comparison, I ran it on my laptop a few years ago, a million places took 3.5 hours. My Raspberry Pi required 44 hours.

  • $\begingroup$ As evident from my question as well as other answers here, what I seek is a method to show that $\pi>3.14$. The method should involve as minimal calculations as possible as well as should be based on mathematical ideas which are easy to understand. What you describe in your story is more related to evaluating million digits of $\pi$ and how technology has helped reduce the time needed to get those digits. While I agree with the facts in your story, it does not really give an answer to my question. $\endgroup$
    – Paramanand Singh
    Oct 24, 2017 at 9:38
  • $\begingroup$ My objective was to calculate as many digits as possible but it could also be used for your objective. It is easy to show that the partial sums are alternating over and under estimates. So, you can stop once you find that the lower estimate has exceeded $3.14$. This may not be the most efficient solution but I thought that you wanted one that was easy to understand. I figured this out when I was 16 at school so I think that it qualifies as easy to understand. $\endgroup$
    – badjohn
    Oct 24, 2017 at 10:14
  • $\begingroup$ Yeah it qualifies as very easy to understand and I mentioned the Leibniz series for $\pi/4$ in my question. The trouble is that it takes a reasonably large number of terms to reach desired level of accuracy. $\endgroup$
    – Paramanand Singh
    Oct 24, 2017 at 11:56
  • $\begingroup$ That's the point of my solution. You need far fewer terms, it is linear: twice as many terms gives you twice as many decimal places. I forget now but I think that it was around 1.6 terms per dp, certainly less than 2. So, 4 or 5 terms would probably be enough. You would not even need my induction step to calculate the $n$th derivative, just calculate the first few long hand. $\endgroup$
    – badjohn
    Oct 24, 2017 at 12:07
  • $\begingroup$ From asin(1/2) I get underestimates only and four terms suffice: $$\pi>6\left(\frac{1}{2}+\frac{1}{6·2^3}+\frac{3}{40·2^5}+\frac{5}{112·2^7}\right)= \frac{56327}{17920}>3.141$$ $\endgroup$ Oct 26, 2017 at 14:32

Approach-1: Using basic geometry (This method is inspired by Archimedes method to estimate $\pi$)

Draw a circle of radius 'r' .

Now, draw a regular polygon (say, polygon-1) with '$N_c$' number of sides such that the circle is circuim-cicle to this polygon. Also, draw another regular polygon (say, polygon-2) with '$N_i$'number of sides such that the circle is in-circle to this polygon.

So,perimeter of Polygon-1 is equal to $C_1=2rN_c sin(\frac{\pi}{N_c})$

Also, it can be found that perimeter of Polygon-2 is equal to $C_2=2rN_i tan(\frac{\pi}{N_i})$

If 'C' is the circumference of the circle, then we know that, $C=2r\pi $

So, by using the fact that 'the sum of any two two sides of a triangle is always grater than the third side' , we can easily show that $C_1<C<C_2$ Putting values, we find that, $2rN_c sin(\frac{\pi}{N_c})<2r\pi<2rN_i tan(\frac{\pi}{N_i})$

This simplifies as, $N_c sin(\frac{\pi}{N_c})<\pi<N_i tan(\frac{\pi}{N_i})$ .......(1)

So, for any two assumed positive values of $N_c$ and $N_i$ , the relation (1) is always valid and is satisfied. As we can observe easily, if $N_c$ and $N_i$ gets infinitely larger, then both polygons-1 & 2 as mentioned above approximates the circle of radius 'r'more closely.

Let, $N_c=N_i=N$ , then it can be proved very easily that, $ \lim_{{N}\to{\infty}} Nsin(\frac{\pi}{N})=\pi$ and $ \lim_{{N}\to{\infty}} Ntan(\frac{\pi}{N})=\pi$

So, inequality (1) is the simplest way to get lower and upper bounds for $\pi$ and these bounds can be improved as per our choice of values of$N_c$ and $N_i$ as per above discussion.

For example. $N_c=N_i=N=4$ gives a bound as, $4 sin(\frac{\pi}{4})<\pi<4 tan(\frac{\pi}{4})$ i.e. $2\sqrt{2} <\pi<4 $

Also, if we increase $N$ to $N_c=N_i=N=6$ , we get a bound as, $6 sin(\frac{\pi}{6})<\pi<6 tan(\frac{\pi}{6})$ i.e. $3 <\pi<2\sqrt{3} $ which is a more tighter bound and so on. It can be found numerically that, when N>57, we have $3.14 <\pi $

Also, if $N_c=6 , N_i=4$ then, using the inequality $N_c sin(\frac{\pi}{N_c})<\pi<N_i tan(\frac{\pi}{N_i})$ , we find that $6 sin(\frac{\pi}{6})<\pi<4 tan(\frac{\pi}{4})$ i.e.$ 3<\pi<4 $

So, $\pi$ can be approximated as the average of these limits as $N_c,N_i\to\infty$

Approach-2: Riemann Zeta function:

We know a well known and famous relation that was given by Euler that,

if 'n' is a positive integer and if $\zeta(2n)=\frac{1}{1^{2n}}+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{4^{2n}}+.....$ then, $\zeta(2n)=(-1)^{n+1}\frac{B_{2n}2^{2n-1}\pi^{2n}}{(2n)!}$

Separating $\pi$ from this relation, we get $\pi^{2n}=\frac{(2n)!\zeta(2n)}{(-1)^{n+1}B_{2n}2^{2n-1}}$

So, $\pi=(\frac{(2n)!\zeta(2n)}{(-1)^{n+1}B_{2n}2^{2n-1}})^{\frac{1}{2n}}$

As it can be observed easily that $\lim_{{n}\to{\infty}}\zeta(2n)=1$ Hence it can be concluded that $\pi=\lim_{{n}\to{\infty}}(\frac{(2n)!}{(-1)^{n+1}B_{2n}2^{2n-1}})^{\frac{1}{2n}}$

Since $\sum_{r=1}^{k}\frac{1}{r^{2n}}<\zeta(2n)$ , hence $\pi>(\frac{(2n)!\sum_{r=1}^{k}\frac{1}{r^{2n}}}{(-1)^{n+1}B_{2n}2^{2n-1}})^{\frac{1}{2n}}$

Actual application of this relationship is quite computationally time consuming but it's really a good approximation when 'n'and 'k' are fairly high values (even $n>8$ and $k>10 $ will give a reasonably good and closer higher bound for $\pi$)


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