New series formula for $\arctan(x)$? Ln(x)? 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.
 A: Let's make it look nice.
∑[(x^(2n+1))/((x^2+1)^(n+1))]*[(2n!!)/(2n+1)!!] 
You say
$\arctan(x)
=\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}\dfrac{(2n)!!}{(2n+1)!!} 
$
Since
$(2n+1)!!
=\prod_{k=1}^n (2k+1)
=\dfrac{\prod_{k=1}^n (2k)(2k+1)}{\prod_{k=1}^n (2k)}
=\dfrac{(2n+1)!}{2^nn!}
$
and
$(2n)!!=2^nn!
$,
this becomes
$\arctan(x)
=\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}\dfrac{2^nn!}{\dfrac{(2n+1)!}{2^nn!}} 
=\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}\dfrac{(2^nn!)^2}{(2n+1)!} 
$.
This series seems to be due to Euler
and is in the
Wikipedia article on
"Inverse trigonometric functions"
at the end of the section
on infinite series:
https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Infinite_series
What you have found is,
as expected, not new.
However,
if you found it by yourself,
that is quite impressive.
A: 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.
A: Evidently from marty cohen's answer, the formula for $\arctan$ was previously known, and as shown below, the result for $\log(x)$ is not too hard to derive. That you have discovered them yourself is no less impressive however.

Substituting $x\mapsto4x^2$ in $(2)$ from this answer, we get
$$
\sum_{k=0}^\infty\frac{4^kx^{2k}}{\binom{2k}{k}}=\frac1{1-x^2}\left[1+\frac{x}{\sqrt{1-x^2}}\sin^{-1}(x)\right]\tag{1}
$$
Integrating $(1)$ gives
$$
\begin{align}
\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}}
&=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\\
&=\frac{\tan^{-1}\left(\frac{\large x}{\sqrt{1-x^2}}\right)}{\sqrt{1-x^2}}\tag{2}
\end{align}
$$
Substituting $x\mapsto\frac{x}{\sqrt{1+x^2}}$ in $(2)$ yields
$$
\frac1{\sqrt{1+x^2}}\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^k}
=\sqrt{1+x^2}\tan^{-1}(x)\tag{3}
$$
and therefore,
$$
\bbox[5px,border:2px solid #C0A000]{\sum_{k=0}^\infty\frac{4^kx^{2k+1}}{(2k+1)\binom{2k}{k}\left(1+x^2\right)^{k+1}}
=\tan^{-1}(x)}\tag{4}
$$

$$
\begin{align}
\sum_{n=1}^\infty\frac{x^n-1}{n(1+x)^n}
&=\log\left(1-\frac1{1+x}\right)-\log\left(1-\frac{x}{1+x}\right)\\
&=\log(x)\tag{5}
\end{align}
$$
