Identifying this approximation of pi

Question

I've come across this approximation of pi, which I'm struggling to put a name on. I want to do further research on it but I can't find any evidence of it on the internet. It is an infinite sum: $\pi = 1 + 1/2 + 1/3 + 1/4 - 1/5 + 1/6 + 1/7 + 1/8 + 1/9 - 1/10 + 1/11 + 1/12 - 1/13 + \dots $

This is the description I've been given of it:

In this formula, each fraction 1/n has a sign (±) determined by: the first two terms have positive signs; after that, if the denominator is a prime of the form 4m − 1 (for example, n = 3, 7, 11, . . .), the sign is positive; if the denominator is a prime of the form 4m + 1 (for example, n = 5, 13, 17, . . .), the sign is negative; if the denominator is a composite number, then the sign is equal to the product of the signs corresponding to its factors (for example, n = 9 = 3 × 3, so its sign is positive (a positive times a positive), while n = 10 = 5 × 2, so its sign is negative (a negative times a positive)).

Can anyone help me identify which approximation this is? Is it a well-known approximation?

Answer 1

Not an answer, but maybe this will help. You can write the sum as an infinite product. For primes $p$ that are $2$ or of the form $4n-1$ you have a factor $$1+\frac 1p+\frac 1{p^2}+\frac 1{p^3}+\ldots=\frac 1{p-1}$$ For primes $q$ of the form $4n+1$ you have a factor $$1-\frac 1q+\frac 1{q^2}-\frac 1{q^3}+\ldots=\frac 1{q+1}$$ So the whole sum becomes $$\prod \frac 1{p_i-1} \cdot \prod \frac 1{q_i+1}$$ I don't know how to relate this to $\pi$

Answer 2

The given sum is related to the Dirichlet $L$-series associated to the quadratic character $\chi_1:(\Bbb Z/4)^\times\to\pm 1$.

en.wikipedia.org/wiki/Dirichlet_character

In a table as the one in loc. cit., Dirichlet_character#Modulus_4: $$ \begin{array}{|r||c|c|c|c|} \hline n\text{ mod }4 & 0 & 1 & 2 & 3 \\\hline \chi_1(n) & 1 & 0 & -1 & 0\\\hline \end{array} $$ But there are some differences.

• First, hen considering the prime $2$, we have $\chi_1(2)=0$, so we have to manually insert this Euler factor.
• Then when considering the signs to be implemented. The signs for the primes are "reverted". We use a minus sign for the primes congruent to $1$ modulo four, and a plus sign for the others. Let us use an other letter for this... $$ \begin{array}{|r||c|c|c|c|} \hline n\text{ mod }4 & 0 & 1 & 2 & 3 \\\hline \eta(n) & -1 & 0 & +1 & 0\\\hline \end{array} $$

In the cited link, the value for $L(\chi_1,s)=\beta(s)$ is referenced to be the Dirichlet beta function computed in $s$.

Putting all together, the given sum $S$ is: $$ \begin{aligned} S &= \left(1+\frac 12+\frac 14+\frac 18+\dots\right) \left(\sum_{\substack{n>0\\n\text{ odd}}}\frac{\eta(n)}n\right) \\ &= \frac 1{1-2^{-1}}\cdot \prod_{p\text{ odd prime}}\frac{1}{1-\eta(p)p^{-1}} \\ &= 2\cdot \prod_{p\text{ odd prime}}\frac{1}{1+\chi_1(p)p^{-1}} \\ &= 2\cdot \prod_{p\text{ odd prime}}\frac{1}{1-p^{-2}} \Big/ \prod_{p\text{ odd prime}}\frac{1}{1-\chi_1(p)p^{-1}} \\[2mm] &=2\cdot \frac{\zeta(2)}{(1-2^{-2})^{-1}}\;\Big/\;L(\chi_1,s)|_{s=1} \\ &=\frac{\pi^2}4\Big/\beta(1) \\ &=\frac{\pi^2}4\Big/\left(1-\frac 13+\frac 15-\frac 17+\frac 19-\dots\right) \\ &=\frac{\pi^2}4\Big/\arctan 1 \\ &=\pi\ . \end{aligned} $$