Continued fraction of $π$ using sums of cubes

Question

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?

Answer 1

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: $$\mathtt{3.14221404702232210406367353362166370131484883936217-}\\\mathtt{3.15126273205858662275081482878228893534757749143403-}$$