Calculate $\pi$ By Hand? All over the internet the only hand equation i found was 
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?
 A: By hand, it's relatively easy to use the development of the arctangent, and a Machin-like formula:
Machin:
$$\frac\pi4=4\arctan\frac15-\arctan\frac1{239}$$
Gauss:
$$\frac\pi4=12\arctan\frac1{18}+8\arctan\frac1{57}-5\arctan\frac1{239}$$
I have done it once with Machin's formula and 24 decimals, in a few hours. It's recommended to do it by two methods, to check there is no computation error.
The arctangent is
$$\arctan x=\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{2k+1}$$
Given a number of decimals, you find where to truncate by estimating the rest, and it's easy since it's an alternating series (so the rest is less in absolute value than the first omitted term).
A: Jean-Claude Arbaut has reminded us of the identity
$$
\frac\pi4=4\arctan\frac15-\arctan\frac1{239}.
$$
Let us examine that.  You learned in high school that $\tan\dfrac\pi4=1$, and that
\begin{align}
\tan(\alpha+\beta) & = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \tag 1 \\[10pt]
& =\frac{c+d}{1-cd}
\end{align}
Thus
$$
\arctan c+\arctan d=\alpha+\beta=\arctan\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\arctan\frac{c+d}{1-cd}
$$
From $(1)$ we get
$$
\tan(\alpha+\beta+\gamma+\delta)=\frac{c+d+e+f-cde-cdf-cef-def}{1-cd-ce-cf-de-df-ef+cdef}
$$
where $c,d,e,f$ are the respective tangents of $\alpha,\beta,\gamma,\delta$, and hence
$$
\tan(4\alpha) = \frac{4\tan\alpha-4\tan^3\alpha}{1 - 6\tan^2\alpha+\tan^4\alpha}.
$$
Hence
$$
4\arctan c = \arctan\frac{4c-4c^3}{1-6c^2+c^4}.
$$
So
$$
4\arctan\frac15 = \arctan\frac{(4/5)-(4/5^3)}{1-(6/5^2)+ (1/5^4)} = \arctan\frac{480}{476} = \arctan\frac{120}{119}.
$$
Next we look at
\begin{align}
& 4\arctan\frac15 - \arctan\frac{1}{239} = \arctan\frac{120}{119} - \arctan\frac{1}{239} \\[15pt]
= {} & \arctan\frac{(120/119)-(1/239)}{1+(120/119)(1/239)} \\[15pt]
= {} & \arctan\frac{28561}{28561} = \arctan 1 = \frac\pi4.
\end{align}
A: The fastest known formula for calculating the digits of pi is Chudnovsky formula:
$$\frac{1}{\pi}=12 \sum_{k=0}^\infty \frac{(-1)^k (6k)! (163 \cdot 3344418k + 13591409)}{(3k)! (k!)^3 640320^{3k+1.5}}$$
This formula is used to create world record for the most digits of pi. This formula rapidly converges and it needs 3-4 terms to yield good approximation of pi which is possible by hand.
A: One easy-to-understand improvement to your method, which I I don't see used much, is:
$$\pi/6 = \arctan \left ( \frac{\sqrt{3}}{3} \right ) \\
 = \int_0^{\frac{\sqrt{3}}{3}} \frac{1}{1+x^2} dx \\
= \sum_{n=0}^\infty \frac{(-1)^{n} \left ( \frac{\sqrt{3}}{3} \right )^{2n+1}}{2n+1} \\
= 3^{-1/2} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1}.$$
Consequently we have
$$\pi = 2 \sqrt{3} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1} \\
\approx 2 \sqrt{3} \sum_{n=0}^N \frac{(-1)^{n} 3^{-n}}{2n+1} \\
= 2 \sqrt{3} \left ( 1 - \frac{1}{9} + \frac{1}{45} - \frac{1}{189} + \dots \right )$$
This gives each decimal digit in slightly fewer than $\log_3(10) \approx 2.1$ steps, provided you can accurately estimate $\sqrt{3}$ to do the final multiplication.
A: For some fast-converging ideas, I recommend looking at this part of the wikipedia page.  For example, the first option they present is
$$
\frac{\pi}{2} = \sum_{k=0}^\infty\frac{k!}{(2k+1)!!}=\\
\frac{1}{1} + \frac{1}{3 \cdot 1} + \frac{2\cdot1}{5\cdot3\cdot1} + \frac{3 \cdot2 \cdot 1}{7\cdot 5 \cdot 3 \cdot 1} + \cdots
$$
Taking this out to the eighth step gives us $\pi \approx 3.137$
A: You need a method to calculate exact $\pi$ by hand by an iteration? Or just some good numerical approximations, which easy to evaluate?
In the first case you can have this.
$$ \pi = \left( \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}} \right)^{-1}. $$
This is a formula by Ramanujan and this gives 8 correct decimal digits for each $k$.
A little bit hard to evaluate Ramanujan's $\pi$ approximation by hand, but if you need just good numerical approximation, for $k=0$ you get this. 
$$\frac{9801\sqrt{2}}{4412} \approx \underline{3.141592}729 $$
There are other formulas which are easier to evaluate by hand using just the elementary operations.
$$
3+\frac{4}{28}-\frac{1}{790+\frac56} \approx \underline{3.141592653}921
$$
This one is good for $9$ decimal digits, and it is easy to remember because the formula contains each number from $0$ to $9$ exactly once.
This one is also good for $9$ decimal places.
$$\frac{22}{17} + \frac{37}{47} + \frac{88}{83} \approx \underline{3.141592653}467 $$
This famous one is good for $6$ places, and also easy to remember:
$$\frac{355}{113} \approx \underline{3.141592}920$$
This one is also interesing, the nominator is $2^9$, and the denominator is the largest Heegner number. This is good for $3$ places:
$$\frac{512}{163} \approx \underline{3.141}104$$
The following one is used since at least Archimedes, and this is good for $2$ places:
$$\frac{22}{7} \approx \underline{3.14}285$$
Here is the proof that $22/7$ exceeds $\pi$.
You could find other approximations at Wikipedia and Wolfram MathWorld.
A: I see Ramanujans formula here, but the one by Chudnovsky Brother's is still missing. Lets change that:
$${\pi =1/12\, \left( \sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k} \left( 6\,k \right) !\, \left( 545140134\,k+13591409 \right) \\
\mbox{}}{ \left( 3\,k \right) !\, \left( k! \right) ^{3}{640320}^{3\,k+3/2}}} \right) ^{-1}}$$
If you do it for the first iteration you get 
$${\frac {53360 \sqrt{640320}}{13591409}} \approx \underline{3.1415926535897}342077$$
Pretty neat, but I wouldn't want to do this "by hand".
A: I skimmed the other answers. I believe this one is different in that it gives a way to rapidly calculate the digits of $\pi$ and proves from scratch that that number in fact is $\pi$.
It's obvious that $\pi = 6 \times \sin^{-1}(\frac{1}{2})$. It turns out that the derivative of $\sin^{-1}$ is elementary so we can take its power series centered at 0 and then integrate it to get the power series of $\sin^{-1}$. In general, $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$. So $\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-\frac{1}{2}}$. Once I get the power series of $(1 + x)^{-\frac{1}{2}}$, I can substitute $-x^2$ for $x$ to get the power series of $(1 - x^2)^{-\frac{1}{2}}$. The first derivative of $(1 + x)^{-\frac{1}{2}}$ is $-\frac{1}{2}(1 + x)^{-1\frac{1}{2}}$. The second derivative is  $-\frac{1}{2}(-1\frac{1}{2})(1 + x)^{-2\frac{1}{2}}$ and so on. Now we divide by the factorials to get the coefficients of the power series so $(1 + x)^{-\frac{1}{2}} = 1 + -\frac{1}{2}x + -\frac{1}{2}(-\frac{3}{4})x^2 + -\frac{1}{2}(-\frac{3}{4})(-\frac{5}{6})x^3 ...$ So the power series of $(1 - x^2)^{-\frac{1}{2}}$ is $1 + \frac{1}{2}x^2 + \frac{1}{2}(\frac{3}{4})x^4 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})x^6 ...$ Then the power series of $\sin^{-1}$ is $x + \frac{1}{2}(\frac{1}{3})x^3 + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})x^5 + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})x^7 ...$ Now finally, $\pi = 6 \times \sin^{-1}(\frac{1}{2}) = 6(2^{-1} + \frac{1}{2}(\frac{1}{3})2^{-3} + \frac{1}{2}(\frac{3}{4})(\frac{1}{5})2^{-5} + \frac{1}{2}(\frac{3}{4})(\frac{5}{6})(\frac{1}{7})2^{-7} ...)$.
