The tangent function has on the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ an inverse function called arctangent and denoted $\arctan$; so $$\tan\bigl(\arctan(x)\bigr)=x\tiny.$$ The graph of $\arctan$ looks like this:

It follows from the definition that $\arctan(1)=\frac{\pi}{4}$ and therefore $$\pi=4\arctan(1)\tiny.$$ If we can calculate a good approximation of $\arctan(1)$, then we can find a good approximation of the number $\pi$. We do this by first determining a Taylor series approximation of the arctangent function.

For the derivative of the tangent function holds: $$\frac{\dd \tan(x)}{\dd x}=\frac{1}{\cos^2x}=1+\tan^2(x)\tiny.$$ From the chain rule it follows with $u=\arctan(x)$ that $$\frac{\dd \tan(u)}{\dd u}\cdot \frac{\dd u}{\dd x}=1$$ and therefore $$\bigl(1+\tan^2(u)\bigr)\cdot\frac{\dd \arctan(x)}{\dd x}=1$$ or $$(1+x^2)\cdot\frac{\dd \arctan(x)}{\dd x}=1$$ In other words: $$\frac{\dd \arctan(x)}{\dd x}=\frac{1}{1+x^2}\tiny.$$ The Taylor series of the right-hand side about $x=0$ can easily be calculated: $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}+\cdots=\sum_{k=0}^{\infty}(-1)^k\cdot x^{2k}\tiny.$$ But then we can also find a series whose derivative is equal to the above Taylor series; this is per construction the Taylor series of $\arctan(x)$ about $x=0$. We get: $$\arctan(x) = x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\frac{1}{9}x^9-\frac{1}{11}x^{11}+\cdots=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\cdot x^{2k+1}\tiny.$$