Continued fraction $[0;2,6,10,14,...,2(2n-1)] = \frac{e-1}{e+1}$

Question

I would like to ask about the following relation, I wonder how to reach it. \begin{equation} K_{n=1}^{\infty} \frac{1}{2(2n-1)} = \frac{1}{2+\frac{1}{6+\frac{1}{10+\cdots}}} = \frac{e-1}{e+1} \approx 0.46 \end{equation} Thank you in advance

Answer 1

One way you may derrive this, is noting that $\text{tanh}(z)=\cfrac{e^{2z}-1}{e^{2z}+1}$. Thus we find that $\text{tanh}(\frac{1}{2})=\frac{e-1}{e+1}$.

And that the non-simple continued fraction of $\text{tanh}(z)$ is given by: $$\text{tanh}(z)=\frac{z}{1+\frac{z^2}{3+\frac{z^2}{5+\cdots}}}$$ From this you find that $$\cfrac{e^{1}-1}{e^{1}+1}=\text{tanh}(\frac{1}{2})=\frac{\frac{1}{2}}{1+\frac{\frac{1}{4}}{3+\frac{\frac{1}{4}}{5+\cdots}}}=\frac{1}{2+\frac{\frac{1}{2}}{3+\frac{\frac{1}{4}}{5+\cdots}}}=\frac{1}{2+\frac{1}{6+\frac{\frac{1}{2}}{5+\cdots}}}=[0,2,6,10,14,\cdots,2(2n-1)]$$