1. Define $\alpha_n = \text{arccot}(\sqrt{2n-1})$. By definition, $\cot(\alpha_n) = \sqrt{2n-1}$.
2. Recall the trigonometric identity for $\cos^2(x)$:
$$\cos^2(x) = \frac{1}{1 + \tan^2(x)}$$
3. Using the double-angle formula for cosine, we have:
$$\cos(2\alpha_n) = 2\cos^2(\alpha_n) - 1$$
4. Substitute $\cos^2(\alpha_n)$ using the identity:
$$\cos^2(\alpha_n) = \frac{1}{1 + \tan^2(\alpha_n)}$$
5. Since $\tan(\alpha_n) = \frac{1}{\cot(\alpha_n)}$, we get:
$$\tan(\alpha_n) = \frac{1}{\sqrt{2n-1}}$$
6. Substitute $\tan(\alpha_n)$ into the expression for $\cos^2(\alpha_n)$:
$$\cos^2(\alpha_n) = \frac{1}{1 + \left(\frac{1}{\sqrt{2n-1}}\right)^2} = \frac{1}{1 + \frac{1}{2n-1}} = \frac{1}{\frac{2n-1+1}{2n-1}} = \frac{2n-1}{2n}$$
7. Now, substitute $\cos^2(\alpha_n)$ back into the double-angle formula:
$$\cos(2\alpha_n) = 2 \left(\frac{2n-1}{2n}\right) - 1 = \frac{2(2n-1)}{2n} - 1 = \frac{4n-2}{2n} - 1 = \frac{4n-2 - 2n}{2n} = \frac{2n-2}{2n} = \frac{n-1}{n}$$
8. Therefore, we have:
$$\cos(2\alpha_n) = \frac{n-1}{n}$$
9. This implies:
$$2\alpha_n = \arccos\left(\frac{n-1}{n}\right)$$
10. Finally, we can express $E_n$ as:
$$E_n = \frac{\arccos\left(\frac{n-1}{n}\right)}{\text{arccot}(\sqrt{2n-1})} = \frac{2\alpha_n}{\alpha_n} = 2$$
11. Since $2$ is a natural number, the result follows.

$\blacksquare$