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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:

$$\sum_{n=0}^\infty \lambda_n^3\, \frac{6n+1}{2^{2n}} =\frac{2^2}{\pi}\tag1$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^3\lambda_n^3}\, \frac{6\big(n-\tfrac12\big)+1}{2^{2n}} =\pi^2\tag2$$


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


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


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

with Catalan's constant $G$.

Level 5:

$$\sum_{n=0}^\infty \lambda_n^5\, \frac{20n^2+8n+1}{(-2^2)^n}=\frac{2^3}{\pi^2}\tag5$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^5\lambda_n^5}\, \frac{20\big(n-\tfrac12\big)^2+8\big(n-\tfrac12\big)+1}{(-2^2)^n} =-56\zeta(3)\tag6$$


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

with Apery's constant $\zeta(3)$.

Level 7:

$$\sum_{n=0}^\infty \lambda_n^7\, \frac{84n^3+38n^2+7n+\tfrac12}{2^{6n}} =\frac{2^4}{\pi^3}\tag9$$ $$\sum_{\color{red}{n=1}}^\infty \frac1{n^7\lambda_n^7}\, \frac{84\big(n-\tfrac12\big)^3+38\big(n-\tfrac12\big)^2+7\big(n-\tfrac12\big)+\tfrac12}{2^{6n}} =\frac{\pi^4}2\tag{10}$$


Most of these are scattered throughout the literature in various guises. See, for example, Guillera and Rogers' paper "Ramanujan Series Upside Down" 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, 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 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".)

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    $\begingroup$ A search using Mathematica's integer relations couldn't find analogous formulas for levels $9,11$ with $(\pm 2^k)^{-n}$ for $k<16$. $\endgroup$ – Tito Piezas III Jan 30 '18 at 16:08
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    $\begingroup$ And there are no others for levels $5,\,7$ with $(\pm2^k)^{-n}$ for $k<16$ either. Sigh. $\endgroup$ – Tito Piezas III Jan 31 '18 at 13:37
  • $\begingroup$ Have you tried level 7 the (n-1/2) type with $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^4}$ (corresponding to G)? $\endgroup$ – Wolfgang Jan 31 '18 at 14:17
  • $\begingroup$ @Wolfgang: Yes, for level $7$ and (n-1/2) type, I tried $\pi^4,1/\pi^4,$ and $\beta(4)$. Only $(10)$ popped up. $\endgroup$ – Tito Piezas III Jan 31 '18 at 14:43
  • $\begingroup$ And I guess no luck with $\beta(3)$ for level 5? - Who knows? $\endgroup$ – Wolfgang Jan 31 '18 at 16:29
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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).

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 there are examples of the "upside-down" technique. Another example is in the Addendum of this paper. In this unpublished file there are many examples of the transformation $n \to n + \frac12$.

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