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These continued fractions for $\pi$ were given here,

$$\small \pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}} = \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1} = \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1 $$

$$\small \pi = 3 + \cfrac{1^2} {6+\cfrac{3^2} {6+\cfrac{5^2} {6+\ddots}}} = 3 - \sum_{n=1}^\infty \frac{(-1)^n} {n (n+1) (2n+1)} = 3 + \frac{1}{1\cdot 2\cdot 3} - \frac{1}{2\cdot 3\cdot 5} + \frac{1}{3\cdot 4\cdot 7} - \cdots\tag2 $$

$$\small \pi = \cfrac{4} {1+\cfrac{1^2} {3+\cfrac{2^2} {5+\cfrac{3^2} {7+\ddots}}}} = 4 - 1 + \frac{1}{6} - \frac{1}{34} + \frac {16}{3145} - \frac{4}{4551} + \frac{1}{6601} - \frac{1}{38341} + \cdots\tag3$$

Unfortunately, the third one didn't include a closed-form for the series. (I tried the OEIS using the denominators, but no hits.)

Q. What's the series formula for $(3)$?

Tito Piezas III
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    Why don't you use the Euler formula to find the series yourself? I assume you know of that formula. See "Euler's formula in modern notation" https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula – peter.petrov Mar 08 '16 at 21:38
  • When you look at the prime factorization of the denominators, you feel little hope of a similarly simple formula. –  Mar 08 '16 at 21:39
  • @peter.petrov: I don't think that's what they used. – Tito Piezas III Mar 08 '16 at 21:40
  • @YvesDaoust: Yes, I was in fact looking at the denominator's prime factorizations too. – Tito Piezas III Mar 08 '16 at 21:40
  • Ah, you mean it's another but equivalent series (with same sum). Hm... – peter.petrov Mar 08 '16 at 21:40
  • Are you sure about that? Because then the equals sign between the continued fraction and the series would be incorrect, no? How should we understand it? Normally it's understood as "the n-th convergent of the continued fraction is equal to the n-th partial sum of the series", no? – peter.petrov Mar 08 '16 at 21:44
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    The ratios of successive terms are very roughly $-1/6$. More than probably, no closed formula is known, except of Ramanujan. –  Mar 08 '16 at 21:45
  • @peter.petrov: If they used the Euler formula, the denominators would have smaller primes dividing them. – Tito Piezas III Mar 08 '16 at 21:49

2 Answers2

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The third one should be obtained from $4.1.40$ in A&S p.68 using $z:=ix$ (from Euler I think not sure) :

$$-2\,i\,\log\frac{1+ix}{1-ix} = \cfrac{4x} {1+\cfrac{(1x)^2} {3+\cfrac{(2x)^2} {5+\cfrac{(3x)^2} {7+\ddots}}}} $$ Except that the expansion of the function at $x=1$ is simply your expansion for $(1)$.

Some neat variants :

$$\varphi(x):=\int_0^{\infty}\frac{e^{-t}}{x+t}dt= \cfrac{1} {x+1-\cfrac{1^2} {x+3-\cfrac{2^2} {x+5-\cfrac{3^2} {x+7-\ddots}}}}$$ $$\text{the previous one was better for large $x$...}$$ $$\int_0^{\infty}e^{-t}\left(1+\frac tn\right)^n\,dt=1+ \cfrac{n} {1+\cfrac{1(n-1)} {3+\cfrac{2(n-2)} {5+\cfrac{3(n-3)} {7+\ddots}}}}$$

$$\sum_{k=0}^\infty\frac 2{(x+2k+1)^2}= \cfrac{1} {x+\cfrac{1^4} {3x+\cfrac{2^4} {5x+\cfrac{3^4} {7x+\ddots}}}}$$ $$\text{and thus $\dfrac{\zeta(2)}2$ for $x=1$ (Stieltjes)}$$

$$\text{The last one was obtained after division by $n$ at the limit $n=0$ :}$$ $$\begin{align} \int_0^1\frac{t^{x-n}-t^{x+n}}{1-t^2}dx&=\sum_{k=0}^\infty\frac 1{x-n+2k+1}-\frac 1{x+n+2k+1}\\ &=\cfrac{n} {x+\cfrac{1^2(1^2-n^2)} {3x+\cfrac{2^2(2^2-n^2)} {5x+\cfrac{3^2(3^2-n^2)} {7x+\ddots}}}}\\ \end{align}$$

Your continued fraction appears too in a neat and recent book by Borwein, van der Poorten, Shallit, Zudilin "Neverending Fractions: An Introduction to Continued Fractions" at the end of the pages $167-169$ reproduced for convenience here (hoping there is no problem with that...) :

p167 p168 p169

Raymond Manzoni
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  • Hm, interesting. If we let $x=iy$ in your answer, what we get is, $$-2\log\frac{1-y}{1+y} = \cfrac{4y} {1-\cfrac{(1y)^2} {3-\cfrac{(2y)^2} {5-\cfrac{(3y)^2} {7-\ddots}}}}$$ What I know is the similar cfrac, $$2\log\frac{y+1}{y-1} = \cfrac{4} {y-\cfrac{1^2} {3y-\cfrac{2^2} {5y-\cfrac{3^2} {7y-\ddots}}}} $$ A third variant is given in this wikipedia article, though it involves squares of odd numbers. – Tito Piezas III Mar 08 '16 at 23:27
  • Those variants can be explained by the cfrac found here

    $$\tan^{-1}\left(\frac{z}{k}\right)=\cfrac{z}{k+\cfrac{(z)^2}{3k+\cfrac{(2z)^2}{5k+\cfrac{(3z)^2}{7k+\cfrac{(4z)^2}{9k+\ddots}}}}}$$

     – Nicco May 17 '16 at 23:34
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Given the symmetric continued fraction found in this post

$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}$$

Your continued fraction $(3)$ is a special case when $a=b=1$.

Moreover ,it has a beautiful q-analogue

$$\begin{aligned}\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}\end{aligned}$$ found here and here.

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Nicco
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