In this Wikipedia article the constant $\pi$ is represented by the following infinite series: $$\pi=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+ \ldots$$

After the first two terms, the signs are determined as follows: If the denominator is a prime of the form $4m−1$, the sign is positive; if the denominator is a prime of the form $4m+1$, the sign is negative; for composite numbers, the sign is equal the product of the signs of its factors.

Now, let us define the function $\operatorname{sgn_1}(n)$ as follows: $$\operatorname{sgn_1}(n)=\begin{cases} -1 \quad \text{if } n \equiv 5 \pmod{6}\\1 \quad \text{if } n \in \{2,3\} \text{ or } n \equiv 1 \pmod{6}\end{cases}$$ 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, $$e=\displaystyle\sum_{n=1}^{\infty} \frac{\operatorname{sgn_2}(n)}{n}$$

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

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


  • Is this representation of the $e$ already known?
  • If it is known can you provide some reference?
  • If it isn't known can you prove or disprove it?

1 Answer 1


Your sum actually equals $\frac{\pi\sqrt{3}}{2}$, so it's more like $\pi$ all over again, not $e$. To see this, note first that by definition $\mathrm{sgn}_2$ is multiplicative, hence $$ A=\sum_{n=1}^{+\infty}\frac{\mathrm{sgn}_2(n)}{n}=\prod_p\left(1-\frac{\mathrm{sgn}_2(p)}{p}\right)^{-1}= $$ $$ =\left(1-\frac12\right)^{-1}\left(1-\frac13\right)^{-1}\prod_{p>3}\left(1-\frac{\mathrm{sgn}_1(p)}{p}\right)^{-1}=3B. $$ Next, if you restrict $\mathrm{sgn_1}$ to $n$ with $(n,6)=1$ (and set it to $0$ elsewhere), you will get the quadratic character $\mod 6$, which is once again multiplicative. Therefore, $$ \prod_{p>3}\left(1-\frac{\mathrm{sgn}_1(p)}{p}\right)^{-1}=B=1-\sum_{n=1}^{+\infty}\left(\frac{1}{6n-1}-\frac{1}{6n+1}\right). $$ The last sum can be transformed into $$ B=1-\sum_{n=1}^{+\infty}\frac{2}{36n^2-1}=\frac{1}{36}\left(36+\sum_{n=1}^{+\infty}\frac{2}{\frac{1}{36}-n^2}\right). $$ From a well-known identity $$ \frac{\pi \cot\pi x}{x}=\frac{1}{x^2}+\sum_{n=1}^{+\infty}\frac{2}{x^2-n^2} $$ we deduce that $$ B=\frac{1}{36}\cdot6\pi\cot \frac{\pi}{6}=\pi\frac{\sqrt{3}}{6} $$ and $$ A=\frac{\pi\sqrt{3}}{2}. $$ One can also notice that $$ \frac{\pi\sqrt{3}}{2}-e\approx 0.0024, $$ so no, these two constants are not even that close.

As for the reference for more general series of this type, see H. Davenport "Multiplicative number theory", Chapter 6 "Dirichlet's class number formula" or Wikipedia page on the same formula


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