# The constant $e$ represented by an infinite series

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?

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.