Define,
$$\lambda\_n =\frac{(\tfrac12)\_n}{(1)\_n} =\frac{(\tfrac12)\_n}{n!} =\frac{\tbinom{2n}{n}}{2^{2n}} =\binom{n-\tfrac12}{n}$$


with *Pochhammer symbol* $(x)\_n$ and *binomial* $\tbinom{n}{k}$. I noticed that the following 14 formulas have a nice "affinity".



> 
> **Level 3:**
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$$\sum\_{n=0}^\infty \lambda\_n^3\, \frac{6n+1}{2^{2n}} =\frac{2^2}{\pi}\tag1$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^3\lambda\_n^3}\, \frac{6\big(n-\tfrac12\big)+1}{2^{2n}} =\pi^2\tag2$$




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$$\sum\_{n=0}^\infty \lambda\_n^3\, \frac{42n+5}{2^{6n}} =\frac{2^4}{\pi}\tag3$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^3\lambda\_n^3}\, \frac{42\big(n-\tfrac12\big)+5}{2^{6n}} =\frac{\pi^2}3\tag4$$




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$$\sum\_{n=0}^\infty \lambda\_n^3\, \frac{4n+1}{(-1)^{n}} =\frac{2}{\pi}\tag i$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^3\lambda\_n^3}\, \frac{4\big(n-\tfrac12\big)+1}{(-1)^n} =-16G\tag{ii}$$




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$$\sum\_{n=0}^\infty \lambda\_n^3\, \frac{6n+1}{(-2^3)^{n}} =\frac{2\sqrt2}{\pi}\tag{iii}$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^3\lambda\_n^3}\, \frac{6\big(n-\tfrac12\big)+1}{(-2^3)^n} =-4G\tag{iv}$$


with *[Catalan's constant](https://en.wikipedia.org/wiki/Catalan%27s_constant)* $G$.



> 
> **Level 5:**
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$$\sum\_{n=0}^\infty \lambda\_n^5\, \frac{20n^2+8n+1}{(-2^2)^n}=\frac{2^3}{\pi^2}\tag5$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^5\lambda\_n^5}\, \frac{20\big(n-\tfrac12\big)^2+8\big(n-\tfrac12\big)+1}{(-2^2)^n} =-56\zeta(3)\tag6$$




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$$\sum\_{n=0}^\infty \lambda\_n^5\, \frac{205n^2+45n+\tfrac{13}4}{(-2^{10})^n}=\frac{2^5}{\pi^2}\tag7$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^5\lambda\_n^5}\, \frac{205\big(n-\tfrac12\big)^2+45\big(n-\tfrac12\big)+\tfrac{13}4}{(-2^{10})^n} =-2\zeta(3)\tag8$$


with *[Apery's constant](https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant)* $\zeta(3)$.



> 
> **Level 7:**
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$$\sum\_{n=0}^\infty \lambda\_n^7\, \frac{84n^3+38n^2+7n+\tfrac12}{2^{6n}} =\frac{2^4}{\pi^3}\tag9$$
$$\sum\_{\color{red}{n=1}}^\infty \frac1{n^7\lambda\_n^7}\, \frac{84\big(n-\tfrac12\big)^3+38\big(n-\tfrac12\big)^2+7\big(n-\tfrac12\big)+\tfrac12}{2^{6n}} =\frac{\pi^4}2\tag{10}$$




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Most of these are scattered throughout the literature in various guises. See, for example, Guillera and Rogers' paper "*[Ramanujan Series Upside Down](https://arxiv.org/pdf/1206.3981.pdf)*" which focuses on level 3. The level 3 formulas for 1/pi were found by Ramanujan and can be explained by modular forms, while $(9)$ is by Gourevitch and $(10)$, in a different guise, is by MO user **zy\_**. In [this post](https://mathoverflow.net/questions/281009/a-mysterious-connection-between-ramanujan-type-formulas-for-1-pik-and-hyperg), he remarked that Guillera, in private correspondence, considered it as new. (Note that its partner was found by Gourevitch and included way back in a [2003 paper](https://projecteuclid.org/download/pdf_1/euclid.em/1087568026) by Guillera.)



> 
> **Q:** What is the unifying theory for these ten formulas, and can we find ***paired*** examples for higher levels, like for $\zeta(5)$? (There is a Ramanujan-type formula for $\zeta(5)$ found by **zy\_** in the post cited, but it does not use $\lambda\_n$ and doesn't seem to have a "partner".)
> 
> 
> 



On a pattern for upside-down Ramanujan pi formulas
All these pairs of formulas are examples of the transformation $n \to n + \frac12$ removing a factor which does not depend on n. The upside-down transformation is essentially $n \to -n$ (therefore it changes $z$ to $z^{-1}$) reinterpreting $(a)\_{-n}$ as $\frac{(-1)^n}{(1-a)\_n}$ if $a \neq 1$, and $(1)\_{-n}$ as $\frac{n(-1)^n}{(1)\_n}$ which preserves formally the recurrence $\Gamma(x+1)=x \, \Gamma(x)$ (see Chapter 7 of the book *A=B* by Petkovsek, Wilf, Zeilberger) and another application to the WZ-method in the Section 4 of [this paper](https://arxiv.org/pdf/1012.2681.pdf)).


The very nice formula for $\zeta(5)$ discovered by zy\_ allows us to discover a new "divergent" (convergent by analytic continuation) Ramanujan-like series for $1/\pi^4$ by using the upside-down-transformation. 


The transformation $n \to n+\frac12$ applied to $\lambda(n)$ essentially inverts $\lambda(n)$ giving $\frac{1}{\lambda(n)}$ but does not invert $z^n$. Hence it is not an upside-down transformation. This explains why the pattern observed in the post cannot be generalized in the way pointed out by the author. 


In the Appendix of [this paper](https://arxiv.org/pdf/1610.04839.pdf) there are examples of the "upside-down" technique. Another example is in the Addendum of [this paper](https://arxiv.org/pdf/1501.06413.pdf). In [this unpublished file](https://drive.google.com/file/d/0B_wpzKF9_poyMU5jWFZfbHlsZjA/view) there are many examples of the transformation $n \to n + \frac12$. 


