Ramanujan's series $1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}$

Solution 1:

Considering $$f(x) =1+\sum_{n=1}^{\infty}\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^{n}$$ $$\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}=\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right) \Gamma (n+1)}$$ and, thanks to a CAS, $$f(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)$$ By the way,

$$g(x)=1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^n$$ write $$g(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)+\frac{x} {32} \, _4F_3\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2;x\right)$$

Edit

Just out of curioisity, considering $$a_n=\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right) \Gamma (n+1)}$$ I had a look at functions $$f_k(x)=1+\sum_{n=1}^{\infty} a_n^k\, x^n$$ and their derivatives and found (probably trivial) the following $$f_2(x)=\, _2F_1\left(\frac{1}{4},\frac{1}{4};1;x\right)$$ $$f_3(x)=\, _3F_2\left(\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1;x\right)$$ $$f_4(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)$$ $$f_5(x)=\, _5F_4\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1,1;x \right)$$ and so on. Similarly $$f_2'(x)=\frac{1}{16} \, _2F_1\left(\frac{5}{4},\frac{5}{4};2;x\right)$$ $$f_3'(x)=\frac{1}{64} \, _3F_2\left(\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2;x\right)$$ $$f_4'(x)=\frac{1}{256} \, _4F_3\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2;x\right)$$ $$f_5'(x)=\frac{1}{1024}\, _5F_4\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2,2;x \right)$$

For $x=1$ $$f_2(1)=\frac{\Gamma \left(\frac{1}{4}\right)}{\sqrt{2 \pi } \Gamma \left(\frac{3}{4}\right)}$$ $$f_3(1)=\frac{\sqrt{\pi }}{\sqrt[4]{2} \Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{7}{8}\right)^2}$$ but, for sure, I have not be able to identify the next terms.

Solution 2:

(Too long for a comment. But this factoid might be useful.) If I remember correctly, a pair of series in that letter was,

$$U_1 = 1-5\left(\frac{1}{2}\right)^{3}+9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}+\cdots=\dfrac{2}{\pi}$$

$$V_1= 1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqrt{2}}{\sqrt{\pi}\,\Gamma^{2}\left(\dfrac{3}{4}\right)}$$ Their similar form can be enhanced as, $$\begin{aligned}U_1&=\sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\V_1&=\sum_{n=0}^\infty (8n+1)\left(\frac{\Gamma\big(n+\tfrac{1}{4}\big)}{n!\;\Gamma\big(\tfrac{1}{4}\big)}\right)^4\end{aligned}$$

$U_1$ belongs to an infinite family,

$$U_1=\sum_{n=0}^\infty\,(-1)^n \left(\frac{(2n)!}{n!^2}\right)^3 \color{blue}{\frac{4n+1}{2^{6n}}}=\frac{2}{\pi}$$ $$U_2=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 \color{blue}{\frac{42n+5}{2^{12n}}}=\frac{16}{\pi}$$

and so on. $V_1$ may also then belong to an infinite family.

To Jack: As relations were asked, maybe the one below will help? Given the binomial $\binom nk$, then we have,

$$\binom{-\tfrac14}{n}\binom{-\tfrac34}{n} = \binom{-\tfrac34}{n}\frac{(-1)^n\,\Gamma\big(n+\tfrac{1}{4}\big)}{\Gamma\big(\tfrac{1}{4}\big)\Gamma(n+1)} = \frac{\binom{4n}{2n}\binom{2n}{n}}{64^n}$$