Is the equation $\pi=\sum_{n=0}^{\infty}\frac{6(-1)^n}{(2n+1)3^{n+1/2}}$ accurate, if so to how many digits? Is it equivalent to other equations? In integrating the unit circle it was found that 
$$\pi=\sum_{n=0}^{\infty}\frac{6(-1)^n} {(2n+1)3^{n+1/2}}$$
The equation was tried in Wolfram Alpha and up to 20 decimal places it seem accurate. The question is, is the equation a true representation of $\pi$? Student's ask, then who or what defines the final value of $\pi$.
 A: Yes, the series is correct in that it is an exact identity.  It is a consequence of the arctangent series $$\tan^{-1} z = \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{2n+1},$$ which is easily derived using Taylor expansion.  Setting $z = 1/\sqrt{3}$ gives the RHS equal to $1/6^{\rm th}$ the given series, and since $\tan^{-1} 1/\sqrt{3} = \pi/6$, the result follows.
A: Too long for a comment.
heropup's answer is a proof that this is correct. Now, the interesting question could be : how many terms have to be summed in order to have $k$ exact digits of $\pi$ in $$\sum_{n=0}^{m}\frac{6(-1)^n} {(2n+1)3^{n+1/2}}$$ Let $S_m$ be the sum of the first $m$
terms of the series. The series is alternating
and and decreasing, so the $m^{\text{th}}$  remainder is such that $$| \pi-S_m | <\frac{6}{(2m+3)3^{m+3/2}}=f_m$$ and we want that $$f_m<10^{-k}$$ Using Lambert function, you can show that $$2m+3 =\frac{2 W\left(3\ 10^k \log (3)\right)}{\log (3)}$$  For large values of the argument, the Wikipedia page reports the approximation
$$W(z)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\cdots$$ where $L_1=\log(z)$ and $L_2=\log(L_1)$. Using the first term only, we would then get $$2m+3\approx 4.19181 k+2.17121 \implies m=2.0959 k-0.414394\approx 2.1 k$$ For $100$ exact digits of $\pi$, this approximation gives $m=210$ while the exact solution would be $m=205$.
For illustration purpose, a few values of $f_m$ are reported below 
$$\left(
\begin{array}{cc}
m & f_m \\
200 & 1.08 \times 10^{-98} \\
201 & 3.58 \times 10^{-99} \\
202 & 1.19 \times 10^{-99} \\
203 & 3.94 \times 10^{-100} \\
204 & 1.31 \times 10^{-100} \\
205 & 4.33 \times 10^{-101} \\
206 & 1.44 \times 10^{-101} \\
207 & 4.77 \times 10^{-102} \\
208 & 1.58 \times 10^{-102} \\
209 & 5.25 \times 10^{-103} \\
210 & 1.74 \times 10^{-103}
\end{array}
\right)$$
As you can notice, it is not a very fast way to compute $\pi$ with many digits.
A: Just for fun $$\sum_{n\geq0}\frac{\left(-1\right)^{n}}{\left(2n+1\right)3^{n}}=\sum_{n\geq1}\frac{\left(-1\right)^{n}}{3^{n}}\int_{0}^{1}x^{2n}dx
 $$ $$=\int_{0}^{1}\sum_{n\geq0}\left(\frac{-x^{2}}{3}\right)^{n}dx=\int_{0}^{1}\frac{3}{x^{2}+3}dx
 $$ $$\overset{x/\sqrt{3}=y}{=}\sqrt{3}\int_{0}^{1/\sqrt{3}}\frac{1}{y^{2}+1}dx=\frac{\pi}{2\sqrt{3}}$$ and so the claim.
