How is the period of a simple pendulum growing with increasing amplitude?

For the period T(α) of a simple pendulum with the length L and the amplitude (the initial elongation) α ∈ (0, π), a strictly increasing sequence Tn (α) is constructed such that the relations T 1 ( α ) = 2 L g π − 2 + 1 ϵ ln 1 + ϵ 1 − ϵ + π 4 − 2 3 ϵ 2 , T n + 1 ( α ) = T n ( α ) + 2 L g π w n + 1 2 − 2 2 n + 3 ϵ 2 n + 2 , $$\begin{array}{c} \displaystyle T_1(\alpha)=2\sqrt{\frac{L}{g}}\left[\pi-2+\frac{1}{\epsilon} \ln\left(\frac{1+\epsilon}{1-\epsilon}\right)+\left(\frac{\pi}{4}-\frac{2}{3}\right)\epsilon^2\right],\\ \displaystyle T_{n+1}(\alpha)=T_n(\alpha)+2\sqrt{\frac{L}{g}}\left(\pi w_{n+1}^2 - \frac{2}{2n+3}\right)\epsilon^{2n+2}, \end{array}$$ and 0 < T ( α ) − T n ( α ) T ( α ) < 2 ϵ 2 n + 2 π ( 2 n + 1 ) , $$\begin{array}{} \displaystyle 0 \lt \frac{T(\alpha)-T_n(\alpha)}{T(\alpha)} \lt \frac{2\epsilon^{2n+2}}{\pi(2n+1)}\,, \end{array}$$ holds true, for α ∈ (0, π), n ∈ ℕ, w n := ∏ k = 1 n 2 k − 1 2 k $\begin{array}{} \displaystyle w_n:=\prod_{k=1}^n\frac{2k-1}{2k} \end{array}$ (the nth Wallis’ ratio) and ϵ = sin(α/2).