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

Abstract

Abstract 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 T1(α)=2Lgπ−2+1ϵln1+ϵ1−ϵ+π4−23ϵ2,Tn+1(α)=Tn(α)+2Lgπwn+12−22n+3ϵ2n+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)−Tn(α)T(α)<2ϵ2n+2π(2n+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, wn:=∏k=1n2k−12k$\\begin{array}{} \\displaystyle w_n:=\\prod_{k=1}^n\\frac{2k-1}{2k} \\end{array}$ (the nth Wallis’ ratio) and ϵ = sin(α/2).