Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$?

Can someone point me to a proof, or explain if it's a simple answer?

What I'm looking for is the point where it becomes understood that trigonometric functions and pi can be expressed as series.  A lot of the information I find when looking for that seems to point back to arctan.
The derivative of the arc tangent is
$$\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}.$$

From the formula for geometric series (see for example this answer for a proof) shows that
$$1+y+y^2+y^3+\cdots = \frac{1}{1-y}\qquad\text{if }|y|\lt 1.$$
Plugging in $-x^2$ for $y$, we get that 
$$\begin{align*}
\frac{1}{1+x^2} &amp;= \frac{1}{1-(-x^2)} \\
&amp;= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\
&amp;= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots
\end{align*}$$
provided that $|-x^2| \lt 1$; that is, provided $|x|\lt 1$. All the computations below are done under this hypothesis (see comments at the end).

So we have that:
$$\frac{d}{dx}\arctan(x) = 1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\qquad\text{if }|x|\lt 1$$
Because this is a Taylor series, it can be integrated term by term. That is, up to a constant, we have:
$$\begin{align*}
\arctan(x) &amp;= \int\left(\frac{d}{dx}\arctan (x)\right)\,dx \\
&amp;= \int\left(1 - x^2 + x^4 - x^6 + x^8 - x^{10}+\cdots\right)\,dx\\
&amp;= \int\left(\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\right)\,dx\\
&amp;= \sum_{n=0}^{\infty}\left(\int (-1)^{n}x^{2n}\,dx\right)\\
&amp;= \sum_{n=0}^{\infty}\left((-1)^{n}\int x^{2n}\,dx\right)\\
&amp;= \sum_{n=0}^{\infty}\left((-1)^{n}\frac{x^{2n+1}}{2n+1}\right) + C\\
&amp;= C + \left( x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} +\cdots\right).
\end{align*}$$
Evaluating at $x=0$ gives $0 = \arctan(0) = C$, so we get
$$\arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \frac{x^{11}}{11} + \cdots,\qquad\text{if }|x|\lt 1.$$
the equality you ask about.

Note however that this does not hold for all $x$: it certainly works if $|x|\lt 1$, by the general properties of Taylor series. But the arc tangent is defined for all real numbers. The series we have here is
$$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}.$$
Using the 
Ratio Test, we have that
$$\begin{align*}
\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} &amp;= \lim_{n\to\infty}\frac{\quad\frac{|x|^{2n+3}}{2n+3}\quad}{\frac{|x|^{2n+1}}{2n+1}}\\
&amp;= \lim_{n\to\infty}\frac{(2n+1)|x|^{2n+3}}{(2n+3)|x|^{2n+1}}\\
&amp;= \lim_{n\to\infty}\frac{|x|^2(2n+1)}{2n+3}\\
&amp;= |x|^2\lim_{n\to\infty}\frac{2n+1}{2n+3}\\
&amp;= |x|^2.
\end{align*}$$
By the Ratio Test, the series converges absolutely if $|x|^2\lt 1$ (that is, if $|x|\lt 1$) and diverges if $|x|\gt 1$. At $x=1$ and $x=-1$, the series is known to converge. So the radius of convergence is $1$, and the equality is valid for $x\in [-1,1]$ only (that is, if $|x|\leq 1$; we gained two points in the process). 

However, the arc tangent has a nice property, namely that 
$$\arctan\left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan(x),$$
So, given a value of $x$ with $|x|\gt 1$, you can use this identity to compute $\arctan(x)$ by computing $\arctan(\frac{1}{x})$ instead, and for this argument the series is valid.
You can show that this series converges for $r = 1$ using the alternating series test.  The sequence of the absolute values of the terms of this series is monotonically decreasing and approach zero. Hence the series converges for $r = 1$. 

Also roughly speaking you need ten times as many terms in the partial sums for the approximation of pi to be one decimal place more accurate.  i.e. If if the first 100 terms have about two decimal places of accuracy, then the first 1000 will give you about three decimal places of accuracy.  Thus, the current record of 12 trillion digits would require a summation of about $10^{12000000000000}$  terms with this series.