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I was looking at this Ramanujan phi-Function

Let: $$R(a)=\sum_{k=1}^{\infty}\frac{1}{(ak)^3-ak}\tag1$$ and this paper of the form,

$$\sum_{k=1}^{\infty}\frac{{2k \choose k}}{4^k}\tag2$$ and $$\sum_{k=1}^{\infty}\left[\frac{{2k \choose k}}{4^k}\right]^2\tag 3$$

By combining them together we have

$$S(a)=\sum_{k=1}^{\infty}\frac{{2k \choose k}}{4^k}\cdot \frac{1}{(ak)^3-ak}\tag4$$ and

$$T(a)=\sum_{k=1}^{\infty}\left[\frac{{2k \choose k}}{4^k}\right]^2\cdot \frac{1}{(ak)^3-ak}\tag5$$

The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are

$$S(2)=\sum_{k=1}^{\infty}\frac{{2k \choose k}}{4^k}\cdot \frac{1}{(2k)^3-2k}=\frac{\pi}{4}-\ln(2)\tag6$$ and

$$T(2)=\sum_{k=1}^{\infty}\left[\frac{{2k \choose k}}{4^k}\right]^2\cdot \frac{1}{(2k)^3-2k}=\frac{6G-\pi\ln(4)-1}{\pi}\tag7$$

Where G is the Catalan's constant.

How can we prove these conjectures?

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  • Is it ${2n \choose n}^2 4^{-2n}$ or ${2k \choose k}^2 4^{-2k}$ in the headline? – Andreas Jan 03 '19 at 09:43
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    Why don't you try the methods of the paper? ;) – Jack D'Aurizio Jan 03 '19 at 17:04
  • Please notice that $(2)$ is actually a divergent series, since $\frac{1}{4^n}\binom{2n}{n}\sim\frac{1}{\sqrt{\pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $\frac{1}{4^n}\binom{2n}{n}$ and its square". – Jack D'Aurizio Jan 03 '19 at 17:20
  • $(3)$ is also a divergent series (same reason), while $$\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^3$$ is related to the lemniscate constant (hence to $\Gamma\left(\frac{1}{4}\right)$) by many good reasons; among them, Clausen's formula for the square of particular $\phantom{}_2 F_1$s. – Jack D'Aurizio Jan 03 '19 at 17:32
  • $$\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4$$ is much more obscure: see https://math.stackexchange.com/questions/2506266/looking-for-a-closed-form-for-a-4-f-3-left-ldots-1-right – Jack D'Aurizio Jan 03 '19 at 17:34

1 Answers1

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Let us tackle $(4)$ first. The ordinary generating function for $\left[\frac{1}{4^n}\binom{2n}{n}\right]$ is $\frac{1}{\sqrt{1-x}}$ and $$ \frac{1}{(an)^3-an}=\frac{1}{(an-1)an(an+1)} = \int_{0}^{1} x^{an}\frac{(1-x)^2}{2x^2}\,dx $$ hence $$ S(a) = \int_{0}^{1}\left(\frac{1}{\sqrt{1-x^a}}-1\right)\cdot\frac{(1-x)^2}{2x^2}\,dx $$ and similarly, since the ordinary generating function for $\left[\frac{1}{4^n}\binom{2n}{n}\right]^2$ is $\frac{2}{\pi}K(x)$, $$ T(a) = \frac{2}{\pi}\int_{0}^{1}\left(K(x^a)-\frac{\pi}{2}\right)\frac{(1-x)^2}{2x^2}\,dx. $$ For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.

(6) and (7) are proved in the mentioned paper, not just conjectured.

Jack D'Aurizio
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