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(A related MSE question by P. Singh.) First define,

$$F_k = 1-5\left(\frac{1}{2}\right)^k+9\left(\frac{1\cdot 3}{2\cdot 4}\right)^k-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^k+17\left(\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\right)^k-\cdots$$

Or more concisely,

$$F_k = \sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{\Gamma\big(n+1\big)\;\Gamma\big(\tfrac{1}{2}\big)}\right)^k$$

The gamma quotient is also,

$$ \frac{\Gamma\big(n+\tfrac{1}{2}\big)}{\Gamma\big(n+1\big)\;\Gamma\big(\tfrac{1}{2}\big)} = \frac1{2^{2n}}\frac{(2n)!}{\,n!^2}$$

Then we have the nice closed-forms,

\begin{align} F_1 &= 0\\[4pt] F_2 &= \dfrac{2\sqrt2\,\Gamma\big(\tfrac{1}{2}\big)}{\Gamma^2\big(\tfrac{1}{4}\big)}\\[4pt] F_3 &= \dfrac{2}{\Gamma^2\big(\tfrac{1}{2}\big)}=\dfrac{2}{\pi} \\[4pt] F_4 &= \; \color{red}{??}\\[4pt] F_5 &= \dfrac{2}{\Gamma^4\big(\tfrac{3}{4}\big)} \end{align}

I found $F_2$ empirically. The closed-form of $F_3=\large\frac{2}{\pi}$ was included by Ramanujan in his letter to Hardy, and $F_5$ is also by him. (I presume a version of $F_2$ may be in his Notebooks.)

Alternatively, while the $F_k$ are clearly a sum of two generalized hypergeometric functions $_pF_q$, we learn from this post that they can also be expressed as just one $_pF_q$,

\begin{align} F_1 &= \,_2F_1\big(\tfrac12,\tfrac54;\tfrac14;-1\big)\\[4pt] F_2 &= \,_3F_2\big(\tfrac12,\tfrac12,\tfrac54;\tfrac14,1;-1\big)\\[4pt] F_3 &= \,_4F_3\big(\tfrac12,\tfrac12,\tfrac12,\tfrac54;\tfrac14,1,1;-1\big)\\[4pt] F_4 &= \,_5F_4\big(\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac54;\tfrac14,1,1,1;-1\big)= \; \color{red}{??}\\[4pt] F_5 &= \,_6F_5\big(\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac54;\tfrac14,1,1,1,1;-1\big) \end{align}

and so on.

Question: What is the closed-form of $F_4,\,F_6$ and others, if any?

Tito Piezas III
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  • Maybe you can reduce the hypergeometic closed form here? – Тyma Gaidash Oct 27 '23 at 13:18
  • @ТymaGaidash Wolfram can actually reduce $F_5$, but I don't know why it has trouble with $F_4$. – Tito Piezas III Oct 27 '23 at 13:23
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    @TitoPiezasIII Once again, an interesting question! Do you have a numerical expansion of $F_4$? Might be worth checking out such a representation with an inverse symbolic calculator – Max Muller Oct 27 '23 at 15:44
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    @MaxMuller Thanks. This WA command should give it to 100 digits, It starts as 0.79443356000157023489670877173541913133327484204150... I assumed it was a product of gamma functions and some radicals, but no hits on Mathematica's integer relations so far. – Tito Piezas III Oct 27 '23 at 16:58
  • If the terms were not alternating in sign we can get the value of $F_4$ using Dougall formula: see https://mathoverflow.net/a/292294/15540 – Paramanand Singh Nov 01 '23 at 14:47
  • @ParamanandSingh I have a feeling it may involve a sum and not just a product of gamma functions. In this 2020 post by Infiniticism, it is discussed how $F_5$ in fact is a consequence of the Dougall formula. – Tito Piezas III Nov 02 '23 at 03:51
  • @ParamanandSingh It seems one person down-voted Tomita's two answers to my question in rapid succession, apparently for no good reason. Can you verify if indeed it was just one person? – Tito Piezas III Jan 07 '24 at 05:12
  • @TitoPiezasIII: voting is anonymous to moderators in the sense that they don't have information on who voted on a specific post. However the timings of the down votes does suggest that it was probably some form of targeted voting. Usually votes in quick succession are reversed by the system. But in any case the user Tomita can raise a flag for targeted voting. – Paramanand Singh Jan 07 '24 at 06:31
  • @ParamanandSingh I see. It is sad the community is tainted by a few bad apples. Thanks for the reply though. – Tito Piezas III Jan 07 '24 at 07:28
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    Although not quite the same, we have

    $$\int_0^1K^2dk=\frac{\pi^4}{32}\sum_{n=0}^{\infty}\binom{2n}{n}^6\left(\frac{1+4n}{2^{12n}}\right)$$

    $$\int_0^1kK^4dk=\frac{3\pi^6}{256}\sum_{n=0}^{\infty}\binom{2n}{n}^8\left(\frac{1+4n}{2^{16n}}\right)$$

     – Miracle Invoker Feb 23 '24 at 05:17

0 Answers0