# Series or product for $\pi$ whose terms are not multiplied by a constant

I'm curious if there exists an infinite sum or infinite product for $$\pi$$ which is not:

1) a Bailey–Borwein–Plouffe-type formula;
2) of the form $$\text{constant}\cdot\sum_{n}a_{n}$$ or $$\text{constant}\cdot\prod_{n}a_{n}$$ (except for the case when the constant is $$-1$$ or $$1$$);
3) of the form $$f\left(\sum_{n}a_{n}\right)$$ or $$f\left(\prod_{n}a_{n}\right)$$ (except for $$f(x)=x$$);
4) a sum or product where $$a_{n}$$ is not elementary.

Some examples of "forbidden" sums and products are $$\pi =\color{red}{4}\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1}$$ $$\pi =\color{red}{2}\displaystyle\prod_{n=1}^\infty \dfrac{4n^2}{4n^2-1}$$ $$\pi =\color{red}{2}\displaystyle\sum_{n=0}^\infty \dfrac{n!}{(2n+1)!!}$$ $$\pi =\left(\color{red}{\dfrac{1}{16}}\displaystyle\sum_{n=0}^\infty \dbinom{2n}{n}^3\dfrac{42n+5}{2^{12n}}\right)^{\color{red}{-1}}$$ $$\pi =\displaystyle\sum_{n=0}^\infty \dfrac{1}{16^n}\left(\dfrac{4}{8n+1}-\dfrac{2}{8n+4}-\dfrac{1}{8n+5}-\dfrac{1}{8n+5}\right) \quad \color{red}{\text{(BBP-type)}}$$ $$\pi =\displaystyle\sum_{n=1}^\infty \dfrac{3^{n}-1}{4^n}\color{red}{\zeta (n+1)} \quad \color{red}{\text{(non-elementary sequence)}}.$$ For example, numbers such as $$e$$ or $$\ln 2$$ satisfy the requirements $$1)$$, $$2)$$, $$3)$$ and $$4)$$: $$e=\displaystyle\sum_{n=0}^\infty \dfrac{1}{n!}$$ $$\ln 2=\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}$$ $$\ln 2=\displaystyle\sum_{n=1}^\infty \dfrac{1}{2^n n}.$$ The formulas for $$\pi$$ that meet the requirements are, for example (note the "$$!$$" symbol, since these are nonsensical): $$\pi \overset{!}{=}\displaystyle\sum_{n=1}^\infty \dfrac{2^n}{n^n (1-2n)}$$ $$\pi \overset{!}{=}\displaystyle\prod_{n=1}^\infty \left(1-\dfrac{1}{n^4+1}\right).$$ Hopefully this question is not too vague and you understand what I mean.

Edit: After reading the comments, I must also emphasize that $$\sum_{n}\text{constant}\cdot a_{n}$$ is "forbidden" as well, since it is the "constant multiplication" in the requirement $$2)$$. Obviously, cases like $$e=\dfrac{1}{2}\displaystyle\sum_{n=0}^\infty \dfrac{2}{n!}$$ do satisfy the requirements, since the constants cancel out.

• You can always pull out a factor out of a series... – amsmath Aug 31 '19 at 18:19
• You don't seem to understand. You will have to clarify what you mean by "not constant $\cdot\sum a_n$". I can always pull the constant inside so that it's not of that form anymore. – amsmath Aug 31 '19 at 18:30
• $\pi=\sum\frac{4\cdot(-1)^n}{2n+1}$. No factor in front. – Wojowu Aug 31 '19 at 18:30
• It'd help if you'd add one or more examples of the type of formula you're looking for rather than keep adding to the list of forbidden types of formulas. – Steven Clark Aug 31 '19 at 20:44
• Why? Like, okay, I can understand looking for interesting formulae representing mathematical constants; but why those... requirements. Yeah, to exclude BBP type series or series for $1/\pi$ seems reasonable, But why on earth is Leibnitz Formula unfitting?! I do not see why an extra factor of $1/4$ is that bad. Could you maybe emphasize why you are looking for such specific examples? – mrtaurho Aug 31 '19 at 22:47

$$\pi=3+\sum_1^{\infty}{(-1)^{n+1}4\over(2n+1)^3-(2n+1)}=3+\sum_1^{\infty}{(-1)^{n+1}\over n(n+1)(2n+1)}$$ is Arndt and Haenel 16.10, attributed to Nilakantha, 15th century, with the reference S. Parameswaran, Whish's showroom revisited, The Mathematical Gazette 76 (1992) 28-36. It's also in Borwein and Borwein, Pi and the AGM, page 101.

$$\pi=3+{1\over6+{9\over6+{25\over6+{49\over6+\cdots}}}}$$ is Arndt and Haenel 16.103, attributed to L J Lange, An elegant new continued fraction for $$\pi$$, American Mathematical Monthly 106 (May 1999) 456-458.

$$\pi=\sum_{-\infty}^{\infty}{\sin n\over n}$$ is formula (16.72) in Arndt and Haenel, $$\pi$$ Unleashed. It can be derived from formula 508 in Jolley, Summation of Series. Jolley refers to Bromwich, Introduction to the Theory of Infinite Series, page 356.

Arndt and Haenel 16.73 is (using the sinc notation) $$\pi=\sum_{-\infty}^{\infty}{\rm sinc}^2n$$ This can be derived from 520 in Jolley, for which the reference is Whittaker and Watson, Modern Analysis (1920 edition), page 163.

• I see what's going on here, but it diverges at $n=0$. There should be some limit so that it's $\lim_{n\to 0}\frac{\sin n}{n}$ instead of $\frac{\sin 0}{0}$. $\pi=2\sum_{n=1}^\infty \frac{\sin n}{n}+1$ would be correct. – Tu1 Sep 1 '19 at 8:26
• Fine, but I'd say it's understood that the $n=0$ term is to be understood as the limit. – Gerry Myerson Sep 1 '19 at 9:06
• But it must be mentioned that it's not the standard fraction. Or, one can use $\pi =\sum_{n=-\infty}^\infty \operatorname{sinc}n=2\sum_{n=1}^\infty \operatorname{sinc}n+1$. The function $\operatorname{sinc}z$ is continuous everywhere (unlike $\frac{\sin z}{z}$). – Tu1 Sep 1 '19 at 9:32

The goal is to find a series for $$\pi$$ of the form $$\sum a_n$$, where $$a_n$$ is elementary, not BBP and not multiplied by a constant (excluding $$\pm1$$). But this is problematic since the "multiplication" criterion is ill-defined. All $$a_n$$ can trivially be expressed as a product of $$c \cdot \frac{a_n}{c}$$, so in that sense they are all multiplied by a constant, hence all $$a_n$$ are impermissible, so there is no such series. The underlying issue is what remainder is permissible after factoring out the constant. You seem to be implicitly assuming that the remainder must be something like $$n^2+1$$ or $$(-1)^n$$, but without other criteria, these are no more natural than the remainder $$\frac{1}{2(n!)}$$ when $$2$$ is factored from $$\frac{1}{n!}$$.

If you just want $$a_n$$ that can simply be expressed with no constant factor, the series

$$\pi=\sum_{n=0}^\infty\frac{50n-6}{2^n\binom{3n}{n}}$$

suffices (Bailey 2019). As with any other, these terms can be expressed as a multiple of something, since they equal $$2\cdot\frac{25n-3}{2^n\binom{3n}{n}}$$. Alternatively, you can specify the remainders to be integers and look for a series $$\sum\frac{b_n}{c_n}$$, such that no constant (except $$\pm1$$) divides all $$b_n$$ or $$c_n$$. For this, you can take Gerry Myerson's series or the series

$$\pi=\sum_{n=1}^\infty\frac{(-1)^{f(n)}}{n}$$

where $$f(n)=0$$, if $$n$$ is a prime of the form $$4k − 1$$; $$f(n)=1$$, if $$n$$ is a prime of the form $$4k + 1$$; $$f(n)$$ is the sum of $$f(i)$$ for the factors, $$i$$ of $$n$$, if $$n$$ is composite (Euler 1748). Whether there's an elementary series that fits the criteria, I'm not yet sure.