In [this Wikipedia article](https://en.wikipedia.org/wiki/List_of_formulae_involving_%CF%80#Infinite_series) 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](https://sagecell.sagemath.org/?z=eJw1jc0KwjAMgO97il6EROpPJnopeRKZ4HSbhZqVrj05fXZrmTl9gXxfUsvHfR5TTYMQCPLL9iBqpU7MpOZZCXOtFjjo0MUUBAj_tCFE8_7ZdbEd99dbHENejGcfxjtYJu06GeID3Ad1eeTOVlODlwJ1kxtLz5caT-lZvNTqkra4BtrZfFf5YCXCtKYtmi9zQDmP&lang=gp&interacts=eJyLjgUAARUAuQ==).


**Question:**


* 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?



The constant $e$ represented by an infinite series
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](https://en.wikipedia.org/wiki/Class_number_formula)


