familiar form?

Relevant wiki: Maclaurin Series

Using Maclaurin series, we have:

$\begin{aligned} \sum_{k=0}^\infty \frac {(-1)^k}{2k+1}x^{2k+1} & = \tan^{-1} x &\small \color{#3D99F6} \text{Putting }x = 1 \\ \sum_{k=0}^\infty \frac {(-1)^k}{2k+1} & = \tan^{-1} 1 = \frac \pi 4 \\ 4 \sum_{k=0}^\infty \frac {(-1)^k}{2k+1} & = \pi \approx \boxed{3.142} \end{aligned}$

this is the idea: $4\displaystyle \sum_{i = 0}^{\infty}\dfrac {(-1)^k}{2k+1} = \pi$ and $\pi = 3.1415...$

i get this from wolfram alpha

i'm sorry for who doesn't know this exspression