TITLE: New series formula for $\arctan(x)$? Ln(x)?
QUESTION [6 upvotes]: I discovered this equation, but have no idea if it has been previously discovered.  Please help determine if it has been previously developed. Or please prove that the equation is not correct.
$$\sum_{n=0}^\infty \frac{x^{2n+1}}{(x^2+1)^{n+1}}\cdot\frac{(2n)!!}{(2n+1)!!}=\arctan(x),$$
for $|x|\leq \pi$, or possibly all $x$.
Likewise, using the same method
for $x> .001$, or possibly  x > 0.
$$\sum_{n=1}^\infty \frac{x^{n}-1}{(1+x)^{n}}\cdot\frac{(1)}{(n)}=Ln(x),$$
all follows from dx/dx =1.

REPLY [3 votes]: The first series is indeed the Euler's transform of the classical Taylor series. From $$f\left(x\right)=\frac{\arctan\left(\sqrt{x}\right)}{\sqrt{x}}=\sum_{n\geq0}\frac{\left(-1\right)^{n}}{2n+1}x^{n}
 $$ we have that $$\frac{1}{1-x}f\left(\frac{x}{x-1}\right)=\sum_{n\geq0}\left(\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k}}{2k+1}\right)x^{n}
 $$ and the sum inside the parentheses is quite simple to evaluate $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k}}{2k+1}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\int_{0}^{1}x^{2k}dx
 $$ $$=\int_{0}^{1}\left(1-x^{2}\right)^{n}dx=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}
 $$ hence $$\arctan\left(x\right)=\color{red}{\sum_{n\geq0}\frac{\left(2n\right)!!}{\left(2n+1\right)!!}\frac{x^{2n+1}}{\left(1+x^{2}\right)^{n+1}}}
 $$ as wanted. The second series is well and easily explained by robjohn.