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.

**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?