# 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.

• You had written "Abs(PI)<= x". I've changed it according to your comment. – Jonas Meyer Jun 22 '16 at 3:57
• There are many series for arctan, including a particularly nice Taylor series. Supposing this series does converge to arctan for some $x$--something you have to prove, btw, not that we have to disprove--what does it matter? – JMJ Jun 22 '16 at 4:00
• apparently, $\displaystyle\arctan(x) = \sum_n \frac{2^{2n} (n!)^2}{(2n+1)!} \frac{x^{2n+1}}{(1+x^2)^{n+1}}$ is the Euler transform of $\displaystyle\arctan(x) = \int \frac{dx}{1+x^2} = \int \sum_{n=0}^\infty (-1)^n x^{2n} dx = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$ (for $|x| < 1$) – reuns Jun 22 '16 at 4:02
• @AndréNicolas: Thanks--I was thinking totally wrongly about $\tan$ instead of $\arctan$. (No excuse but perhaps the suggestion of $\pi$ being relevant to the domain had thrown me off.) – Jonas Meyer Jun 22 '16 at 4:02
• The error at n = 112 x= PI is -0.00000360894971418987 and the series coverges – 926reals Jun 22 '16 at 4:03

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.

• can you say if it is the Euler transform of $\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$ ? – reuns Jun 22 '16 at 4:06

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.

• Nor Euler's transform nor Taylor series needed. Jut the definition of the derivative. – 926reals Oct 29 '16 at 4:42
• @926reals What? I proved that your series is the Euler transform of the Taylor series. So I don't understand your comment. – Marco Cantarini Oct 29 '16 at 6:51
• Nothing wrong with your reply. Absolutely nothing. It is best I seen. But I asked the wrong question. I should have ask, how to derive Tan(x) and LN(x) without use of power series or Euler transform or anything else but just primordial calculus. Based on your feedback I posted the derivations. – 926reals Oct 30 '16 at 3:58

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}

• $(2)$ is also $(10)$ from this answer. – robjohn Oct 24 '16 at 11:34
• @ robjohn. Thank you I see. – 926reals Oct 25 '16 at 2:38