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$$ \mbox{Is this closed form true ?:}\quad \sum_{n \geq 1}{n + 1/\left(4n\right) - 1 \choose n} =\frac37 $$

The series arises upon taking $\lambda=0$ in the identity from another post $$ \pi = 4+\sum_{n=1}^\infty \frac1{n!}\left(\frac1{n+\lambda}-\frac4{2n+1}\right)\left(\frac{(2n+1)^2}{4(n+\lambda)}-n\right)_{n-1}$$ Understanding how to work with fractional binomials of the form $\binom{Q(n)}n$ with $Q$ rational could be an initial step towards proving the $\pi$ identity purely mathematically — the proof currently requires knowledge of quantum physics concepts.

If $Q$ were constant, generating functions would be the natural approach but the presence of the $1/(4n)$ term gets in the way here.

Felix Marin
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TheSimpliFire
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    What's the difference between quantum physics and mathematics anyway? – Oscar Lanzi Jun 27 '24 at 12:52
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    @OscarLanzi I intended to mean that the current proof of the $\pi$ identity requires knowledge of physics. My ultimate goal is to see a proof that doesn't require that. The particular series I'm asking about here is only one of two terms of the $\pi$ identity when $\lambda=0$. – TheSimpliFire Jun 27 '24 at 12:59
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    Nice question.. – GEdgar Jun 27 '24 at 14:00
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    Plugging in $\lambda=0$, what I get is $$\pi=4-4\sum_{n=1}^\infty \binom{n+1/(4n)-1}{n}\frac{2n-1}{2n+1}.$$ So it's not evident to me how you the $3/7$ conjecture from, though I'll grant that the numerical agreement is (at least initially) reasonable. – Semiclassical Jun 27 '24 at 15:12
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    @Semiclassical I got only the series from the $\pi$ identity, not the $3/7$ value. I conjectured the value after splitting $(2n-1)/(2n+1)$ into $\color{red}1-\color{blue}{2/(2n+1)}$ and plugging the first term into Mathematica. – TheSimpliFire Jun 27 '24 at 15:55

1 Answers1

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Compute the series of the summand at infinity to arrive at $$\binom{n+\frac{1}{4n}-1}{n}=\frac{1}{4n^2}+\frac{\gamma+\log(n)}{16n^3}+O\left(\frac{1}{n^4}\right).$$

Consider say $$\frac{\pi^2}{24}+\sum_{n=1}^{\infty}\left(\binom{n+\frac{1}{4n}-1}{n}-\frac{1}{4n^2}\right)$$ which converges faster than our original series. Computing up to $3000$ terms, Mathematica converges to a value of $0.42842046\dots < \frac{3}{7}=0.42857142\dots$, so your conjectured closed form appears to sadly not be true. We can keep subtracting higher order terms in the summand to increase convergence speed.

Alternatively, note that your sum can also be written as $$\sum_{n=0}^{\infty}(-1)^{n-1}\binom{-\frac{1}{4(n+1)}}{n+1}$$ which allows us to apply Euler-type series acceleration methods instead. We can also combine our original method with say Aitken's delta-squared process to further increase convergence speed.

KStar
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