Inspired by the Riemann Hypothesis, I have been looking for different ways to relate the number $\frac{1}{2}$ to some interesting property related to the Harmonic series, and this is the outcome:
Let us consider the harmonic series $$\sum_{n=1}^{\infty}\frac{1}{n}$$
Let us consider also the series $$\sum_{n=1}^{\infty} x^n$$
It is not difficult to prove that, when $x=\frac{1}{2}$, $$\sum_{n=1}^{\infty} x^n=1$$
Hence, when $x=\frac{1}{2}$, $$\sum_{n=1}^{\infty}\frac{1}{n}=\left(\sum_{n=1}^{\infty}\frac{1}{n}\right)\left(\sum_{n=1}^{\infty} x^n\right)$$
Expanding the RHS, we get that
$$\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots\right)\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots\right)=$$
$$\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\dots\right)+\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{16}+\dots\right)+\left(\frac{1}{8}+\frac{1}{16}+\frac{1}{24}+\frac{1}{32}+\dots\right)+\dots=$$
$$\frac{1}{2}+2\left(\frac{1}{4}\right)+\frac{1}{6}+3\left(\frac{1}{8}\right)+\frac{1}{10}+2\left(\frac{1}{12}\right)+\dots$$
It can be appreciated that, amazingly enough, we get a sum of the reciprocals of the even numbers, with each reciprocal weighted by the degree of the power of two that divides it. And thus, setting $n=2^{k}m$, where $m$ is some odd number, we get the identity
$$\sum_{n=1}^{\infty}\frac{1}{n}=\sum_{n=1}^{\infty}\frac{k_n+1}{2n}$$
Note also that this is equal to state that $$2\sum_{n=1}^{\infty}\frac{1}{2n}=\sum_{n=1}^{\infty}\left(k_n+1\right)\frac{1}{2n}$$
Which are the implications of this identity? Can be generalized further in any way? Is this unique, or could there be derived similar identities for other prime numbers?
Thanks in advance for your comments / answers!
EDIT
After looking at the comments, it seems that as $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent, the identity is not valid. Therefore, I would like to check if the process could be applied to the alternating harmonic series $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\ln{2}$$