$$\pi=\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}$$ to $$\pi=\sqrt12(\dfrac{1}{1*3^0}-\dfrac{1}{3*3^1}+\dfrac{1}{5*3^2}-\dfrac{1}{7*3^3}+...)$$
After reading through several books and PDFs regarding series acceleration, I cannot for the life of me actually come to the logical flow necessary to transform the original series into the latter series. I set off several years ago to understand this transformation. Over the years I've learned so much about transformations and can accelerate many series for constants through many glorious methods, however I've never learned how the series that got me started can be derived. I cannot find a direct derivation anywhere.
Can someone please illuminate me?
let $\pi_n=\sum_{i=0}^{n}\frac{(-1)^n}{2n+1} $
$$\pi_n=\sum_{i=0}^{n}\int_0^1(-1)^nx^{2n}dx $$
$$\pi_n=\int_0^1\sum_{i=0}^{n}(-x^2)^ndx $$ $$\pi_n=\int_0^1\frac{1-(-x^2)^{n+1}}{1+x^2} $$ $$\pi_n=\int_0^1\frac{1}{1+x^2} +R_n$$ with $R_n=\int_0^1\frac{(-x^2)^{n+1}}{1+x^2}$. You can see that $|R_n|<\int_0^1x^{2n+2}\leq \frac{1}{2n+3} \rightarrow 0 $ when n goes to infinity. And you have :
$$\pi_n=[arctan(x)]_0^1+R_n=\frac{\pi}{4}+R_n $$
so : $$\pi=4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} $$
you can use the same reasoning for the second sum, (I'll skip some details about the convergence, you can use partial sums as I did above) :
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\sum_{n=0}^{\infty}\int_0^1\left(\frac{-x^2}{3}\right)^n dx$$
$$ \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}= \int_0^1 \frac{1}{1+\frac{x^2}{3}}dx$$ $$ \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}= \sqrt{3}\int_0^{\frac{1}{\sqrt{3}}} \frac{1}{1+y^2}dy$$ $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\sqrt{3}\times arctan(\frac{1}{\sqrt{3}}) $$ $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}=\frac{\pi}{\sqrt{12}} $$
So you get the final result : $$\boxed{ \pi=\sqrt{12}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)3^n}}$$