# Continued fraction of $π$ using sums of cubes

Recently I came across this identity: $$\pi=3+\cfrac1{6+\cfrac{1^3+2^3}{6+\cfrac{1^3+2^3+3^3+4^3}{6+\cfrac{1^3+2^3+3^3+4^3+5^3+6^3}{6+\ddots}}}},\tag1$$ thus $$\pi=3+\cfrac{1}{6+\cfrac{(1\cdot3)^2}{6+\cfrac{(2\cdot5)^2}{6+\cfrac{(3\cdot7)^2}{6+\ddots}}}}.\tag2$$ We also know that $$\pi=3+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n(n+1)(2n+1)},\tag3$$ Converting the sum to a continued fraction we get: $$\pi=3+\cfrac{1}{6\cdot1^2+\cfrac{(1\cdot2\cdot3)^2}{6\cdot2^2+\cfrac{(2\cdot3\cdot5)^2}{6\cdot3^2+\cfrac{(3\cdot4\cdot7)^2}{6\cdot4^2+\ddots}}}}.\tag4$$ Are the two continued fractions $$(2)$$ and $$(4)$$ really equal? And how is the above continued fraction using sums of cubes derived?

• The sum-of-cubes formula was written up by Tony Foster, available at pdfcoffee.com/… . He did algebraic magic starting with numerators that were squares of triangular numbers. Mar 24, 2022 at 16:25
• I should add that I've verified numerically it diverges, as does a couple continued fraction forms earlier on in Foster's derivation. He tried to make a simplified (not "simple") continued fraction after converting the Nilakantha infinite series into a nonsimple continued fraction. Mar 24, 2022 at 17:09

The first two identities $$(1),(2)$$ are wrong; the continued fraction with "sums of cubes" doesn't converge. This can be shown using the known criterion: for a sequence $$\{a_n\}$$ of positive real numbers, the continued fraction $$a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{\ldots}}}$$ converges if and only if the series $$\sum_{n=0}^{\infty}a_n$$ diverges. In our case, for $$n>0$$, we have $$a_n=6c_n$$ where $$c_1=1$$ and $$c_{n+1}=1/\big(n^2(2n+1)^2 c_n\big)$$, and one obtains $$\color{blue}{a_n=\mathcal{O}(1/n^2)}$$ using, say, the $$\Gamma$$-function: $$c_n=d_n\left(\frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{n}{2}+\frac14\right)}{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{n}{2}+\frac34\right)}\right)^2\quad\implies\quad d_{n+1}=\frac{1}{64d_n},$$ giving $$d_n=\mathcal{O}(1)$$ and $$c_n=\mathcal{O}(1/n^2)$$ since $$\Gamma(x+a)/\big(x^a\Gamma(x)\big)\underset{x\to\infty}{\longrightarrow}1$$.
For curiosity, here are computed approximate values of the lower/upper limits, respectively: $$\mathtt{3.14221404702232210406367353362166370131484883936217-}\\\mathtt{3.15126273205858662275081482878228893534757749143403-}$$