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Have you seen this series referenced anywhere?

$$\sum_{n=0}^\infty \frac{(-1)^n}{1-4 n^2} \binom{2n}{n} \frac{\left(x+ \frac{1}{x}-2 \right)^n}{2^{4n}}= \frac{\sqrt{x}}{2} \left(\frac{\log x}{x-1}+\frac{x+1}{2x} \right) \tag{1}$$

It converges for: $$3-2 \sqrt{2} < x < 3+2 \sqrt{2}$$

The convergence rate is better than the well known $\text{arctanh}$ series for small number of terms, but worse for larger number of terms *.

$$\log x=2\sum_{k=0}^\infty\,\frac{1}{2k+1}\left(\frac{x-1}{x+1}\right)^{2k+1}$$

Obviously, there's nothing new about (1), it's just a little more complicated $\text{arcsinh}$ series. However, I think it's very nice, because it highlights the properties of logarithms, including the $x \to 1/x$ antisymmetry, and again, converges fast.

Due to the fact that it's alternating, we can also use a better approximation with the truncated sum:

$$\frac{\sqrt{x}}{2} \left(\frac{\log x}{x-1}+\frac{x+1}{2x} \right) \approx \\ \approx \sum_{n=0}^{N-1} \frac{(-1)^n}{1-4 n^2} \binom{2n}{n} \frac{\left(x+ \frac{1}{x}-2 \right)^n}{2^{4n}}+\frac{(-1)^N}{2(1-4 N^2)} \binom{2N}{N} \frac{\left(x+ \frac{1}{x}-2 \right)^N}{2^{4N}}$$


What's really interesting (to me at least) is how I obtained this series. I didn't even think about hyperbolic functions.

I started with a Frullani integral:

$$\int_0^\infty \frac{1}{t} \left(\frac{1}{1+t}-\frac{1}{1+xt} \right)= \log x$$

Then I used the inverse function identity:

$$y= f(t)=\frac{1}{t} \left(\frac{1}{1+t}-\frac{1}{1+xt} \right)$$

$$\log x= \int_{y(\infty)}^{y(0)} f^{-1} (y) dy$$

We can get the inverse function explicitly:

$$y t (1+t) (1+x t)=(x-1) t$$

$$t^2+\frac{x+1}{x} t+\frac{y-x+1}{x y}=0$$

Solving this simple quadratic and taking the positive root, we obtain:

$$t(y)=\frac{1}{2x} \left( \sqrt{\frac{x-1}{y}} \sqrt{(x-1) y+4x}-(x+1) \right)$$

$$\log x= \int_0^{x-1} t(y) dy$$

Substituting:

$$y=(x-1) z^2$$

We have:

$$t(z)=\frac{1}{2x} \left( \frac{1}{z} \sqrt{(x-1)^2 z^2+4x}-(x+1) \right)$$

$$\log x=\frac{x-1}{x} \int_0^1 \left( \sqrt{(x-1)^2 z^2+4x}-(x+1)z \right) dz$$

Simple rearrangement and expanding the square root immediately leads to (1), including the condition for convergence.

I wonder if other, even more interesting series for the logarithm can be obtained by this method, using the general integral:

$$\int_0^\infty \frac{F(t)-F(x t)}{t} dt=(F(0)-F(\infty)) \log x$$


* Update, more information about convergence rate:

If we denote:

$$f(x)=\frac{\sqrt{x}}{2} \left(\frac{\log x}{x-1}+\frac{x+1}{2x} \right)$$

$$S(x)=\sum_{n=0}^N \frac{(-1)^n}{1-4 n^2} \binom{2n}{n} \frac{\left(x+ \frac{1}{x}-2 \right)^n}{2^{4n}}$$

$$G(x)=\sqrt{x} \left(\frac{1}{x-1} \sum_{k=0}^N \frac{1}{2k+1}\left(\frac{x-1}{x+1}\right)^{2k+1}+\frac{x+1}{4x} \right) $$

Then it's clear that for a very small number of terms $N \leq 10$ and for $x$ close to $1$, $S(x)$ performs better than $G(x)$, but for larger number of terms it's not as good.

enter image description here

enter image description here


There's another series I found for the logarithm, but it has even slower convergence:

$$\log x= \frac{x^2-1}{2x} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1) \binom{2n}{n}} \frac{(x-1)^{2n}}{x^n}$$

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  • $\begingroup$ For which $x$ is the convergence rate better? Based on the problem statement the series presented here loses convergence when $x>3+2\sqrt{2}$ but the arctangent series keeps converging. $\endgroup$ – Oscar Lanzi Jul 18 at 9:58
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    $\begingroup$ @OscarLanzi, you asked the right question, I updated and fixed my wording $\endgroup$ – Yuriy S Jul 19 at 19:37
  • $\begingroup$ Thanks @yuriy, some calculations I did point in a similar direction. I see that the noise in your second comparison plot has all the trappings of roundoff error. $\endgroup$ – Oscar Lanzi Jul 19 at 20:53

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