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I am looking for the proof of the following claim:

First, define the function $\operatorname{sgn_1}(n)$ as follows: $$\operatorname{sgn_1}(n)=\begin{cases} -1 \quad \text{if } n \neq 3 \text{ and } n \equiv 3 \pmod{4}\\1 \quad \text{if } n \in \{2,3\} \text{ or } n \equiv 1 \pmod{4}\end{cases}$$

Let $n=p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot \ldots \cdot p_k^{\alpha_k}$ , where the $p_i$s are the $k$ prime factors of order $\alpha_i$ .

Next, define the function $\operatorname{sgn_2}(n)$ as follows: $$\operatorname{sgn_2}(n)=\displaystyle\prod_{i=1}^k(\operatorname{sgn_1}(p_i))^{\alpha_i}$$

Then, $$\pi=\displaystyle\sum_{n=1}^{\infty} \frac{\operatorname{sgn_2}(n)}{n}$$

The first few terms of this series: $$\pi=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}-\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}+ \ldots$$

The sum of the first $3000000$ terms gives the following result rounded to the $37$ decimal places: $$\displaystyle\sum_{n=1}^{3000000} \frac{\operatorname{sgn_2}(n)}{n}=3.1415836871743792245050824485818285768$$

The SageMath cell that demonstrates this infinite series can be found here.

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    $\begingroup$ Equivalently,$$\pi=3\prod_{p>3}\frac p{p-(-1)^{\frac{p-1}2}}=3\cdot\frac54\frac78\frac{11}{12}\frac{13}{12}\frac{17}{16}\frac{19}{20}\cdots$$(product over primes only). $\endgroup$ Jun 16 at 4:17
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This can be proved similarly as Alexander Kalmynin's method .

Let, the sum be $S$, then we can make the following identity because $\text{sgn}_1$ of $2,3$ is defined to be $1$. So, $\text{sgn}_1(ak)=\text{sgn}_1(k), a=2,3,6$. Also, from the definition of $\text{sgn}_2$ we can see, $\text{sgn}_2(n)=1,-1$ respectively for $n \equiv 1,-1 (\text{modulo} 4)$.

As, $\text{sgn}_1(ak)=\text{sgn}_1(k), a=2,3$, we can separate them out from $S$ in the form of $\frac{S}{2}+\frac{S}{3}$. To prevent double counting of the $6$s multiples, we subtract $\frac{S}{6}$. Then, what is left is all odd numbers which aren't divisible by $3$. So, they can be classified as, $12k+\sigma , \sigma=1,-1,5,-5$ and also these are of the form $4k±1$.

The identity : $$S=\frac{S}{2}+\frac{S}{3}-\frac{S}{6}+\left(\sum_{n\geq 0}\frac{1}{12n+1}-\sum_{n\geq 1}\frac{1}{12n-1}\right)+\left(\sum_{n\geq 0}\frac{1}{12n+5}-\sum_{n\geq 1}\frac{1}{12n-5}\right)$$

This gives $\frac{S}{3}=(1+\frac{1}{5})+\frac{1}{12^2}\left(\sum_{n=1}^{\infty} \frac{2}{(\frac{1}{12})^2-n^2}+\sum_{n=1}^{\infty} \frac{ 10}{(\frac{5}{12})^2-n^2}\right)$

or, $\frac{S}{3}=\frac{\pi}{12}\left(\text{cot}(\frac{\pi}{12})+\text{cot}(\frac{5\pi}{12})\right)$

This gives $S=\pi$

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  • $\begingroup$ Wait, $\operatorname{sgn}_2(15)=1$, although $15\equiv-1\mod4$. Also, $\operatorname{sgn}_2(21)=-1$, although $21\equiv1\mod4$. There are in fact infinitely many counterexamples. $\endgroup$ Jun 16 at 13:33
  • $\begingroup$ Could you please indicate this then? Your statement does not mention this additional restriction. $\endgroup$ Jun 16 at 13:36

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