# An accountancy of the natural numbers

I guess this formula is known since I believe it's true $$\sum_{k=2}^\infty\frac{(-1)^{1+\Omega(k)}}{k}=1$$ where $\Omega(k)$ is the number of prime factors of $k$ (not necessary different primes). I can't find it and want a formal proof or a reference.

The probability of a number to have the prime factor $p$ is $\frac{1}{p}$ and to have the prime factors $p$ or $q$ is $\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}$ etc. The probability of a number $>1$ to have a prime factor is $1$.

Is this sum really not known? Haven't anything been published about it before?

I finally realize that the series $$\sum_{k=1}^\infty\frac{(-1)^{\Omega(k)}}{k}=0$$ has been known for a long time, and that solves my confusion.

• I just want to know how you were able to formulate this conjecture on your own? – BigbearZzz Aug 21 '16 at 23:05
• @BigbearZzz: I have edited my answer about my intuition. – Lehs Aug 22 '16 at 0:02
• Interesting. Here, we have two "simple" proofs. In MO we get the remark that it is quite deep, equivalent to the prime number theorem. mathoverflow.net/questions/248006/an-unknown-series – GEdgar Aug 22 '16 at 16:14

The result is a simple consequence of the properties of Dirichlet series.
Since $f(n)=(-1)^{\Omega(n)}$ is a multiplicative function, for any $s>1$ we have

$$\sum_{n\geq 1}\frac{f(n)}{n^s} = \prod_{p}\left(1-\frac{1}{p^s}+\frac{1}{p^{2s}}-\frac{1}{p^{3s}}+\ldots\right)=\prod_p\left(1+\frac{1}{p^{s}}\right)^{-1}=\frac{\zeta(2s)}{\zeta(s)} \tag{1}$$

by Euler's product. Since $\zeta(s)$ has a simple pole at $s=1$ with residue $1$, by taking the limit as $s\to 1^+$ we get $$\boxed{ \sum_{n\geq 2}\frac{(-1)^{1+\Omega(n)}}{n}=\color{red}{1} }\tag{2}$$ as wanted. As pointed out by Kcd in the comments, we switched a sum and a limit to state the equality above, i.e. we exploited some form of the dominated convergence theorem, that in this case is given by the principle of analytic continuation. So, to the able to state $(2)$ with full rigor, we have to prove first that $$\sum_{n\leq N}(-1)^{\Omega(n)}=o(N) \tag{3}$$ holds. But that is a standard result in sieve theory. For instance, we may go through the following lines: we may choose at first some prime $q$ on the interval $[2,n]$ such that the vast majority of integers in such interval are $q$-smooth. Then the LHS of $(3)$, restricted to such integers, is not too difficult to approximate, and with an accurate choice of $q$ we may simply neglect the contribute given by the other (non-smooth) integers. An alternative is to notice that $(-1)^{\Omega(n)}$ equals $\mu(n)$ if $n$ is a square-free number, square-free numbers have a positive density among integers ($\frac{6}{\pi^2}$) and $\sum_{n\leq N}\mu(n)$ is well studied, so we may try to separate terms appearing in the LHS of $(3)$ according to their square-free part or their largest squarefree divisor and apply double-counting. Luckily, we don't have to be extremely accurate to prove $(3)$, since an almost ridiculous upper bound like $$\sum_{n\leq N}(-1)^{\Omega(n)}\leq\frac{N}{\sqrt{\log\log\log N}}$$ is yet good enough. Moreover, at first sight it looks that both $$\sum_{n\leq N}\mu(n)=o(N),\qquad \sum_{n\leq N}(-1)^{\Omega(n)}=o(N)$$ are equivalent to $\zeta(s)\neq 0$ over $\text{Re}(s)=1$, but I may be wrong about that.

• If a Dirichlet series $\sum_{n \geq 1} a_n/n^s$ converges for ${\text Re}(s) > 1$ and has a limit as $s \rightarrow 1^+$, why does $\sum_{n \geq 1} a_n/n$ converge and equal that limit? This needs justification. For power series you can ask an analogous question: if $\sum_{n \geq 0} c_nz^n$ converges for $|z| < 1$ and has a limit as $z \rightarrow 1^{-}$, must $\sum_{n \geq 0} c_n$ converge and equal that limit? The answer is yes if $c_n \rightarrow 0$ and the power series has an analytic continuation to $z = 1$ (theorem of M. Riesz). Obviously we need $c_n \rightarrow 0$ if (continued) – KCd Aug 23 '16 at 10:55
• the series $\sum_{n \geq 1} c_n$, or more generally $\sum_{n \geq 0} c_nz_0^n$ for some $z_0$ with $|z_0| = 1$, is going to have a chance to converge. Riesz proved an analogue of this result for Dirichlet series: if $(a_1 + \cdots + a_n)/n \rightarrow 0$ and $\sum_{n \geq 1} a_n/n^s$ for ${\text Re}(s) > 1$ has an analytic continuation to $s = 1$ then $\sum_{n \geq 1} a_n/n$ converges. The hypothesis $(a_1 + \cdots + a_n)/n \rightarrow 0$ is "natural" in the sense that it is necessary for $\sum_{n \geq 1} a_n/n$ to converge. (If $\sum_{n \geq 1} a_n/n^{s_0}$ converges and (continued) – KCd Aug 23 '16 at 11:00
• Of course you're right, to exchange the sum and the limit requires some form of the dominated convergence theorem. The situation, here, is quite the same presented by $$\sum_{n\few 1} \frac{\mu(n)}{n}=0$$ that is quite well known. – Jack D'Aurizio Aug 23 '16 at 11:01
• ${\rm Re}(s_0) > 0$ then necessarily $|(a_1 + \cdots + a_n)/n^{s_0}| \rightarrow 0$.) In the case when $a_n = (-1)^{\Omega(n)}$, where $\sum_{n \geq 1} a_n/n^s = \zeta(2s)/\zeta(s)$ for ${\rm Re}(s) > 1$, this function has an analytic continuation to a neighborhood of $s = 1$, so we can justify saying the series at $s = 1$ is the value of $\zeta(2s)/\zeta(s)$ at $s = 1$, namely the value of the series at $s = 1$ is $0$, if it is shown that $\sum_{k \leq n} (-1)^{\Omega(k)} = o(n)$. As for showing that... – KCd Aug 23 '16 at 11:04

We have

$$\sum_{k=2}^{\infty} \frac{(-1)^{1 + \Omega(k)}}{k} = 1 -\prod_p \left(1 - \frac{1}{p} + \frac{1}{p^2} \mp\right) = 1 -\prod_{p} \frac{1}{1 + 1/p}$$

The RHS goes to $1$ since the product goes to zero.

• Rearrangement of a conditionally convergent series? – GEdgar Aug 22 '16 at 13:38
• @GEdgar You're right, the first equality is probably more of a formal power series than anything. I'm not familiar with methods in analytic number theory that would make this rigorous, as Jack does - I just noticed the "identity" and went with it. – MCT Aug 22 '16 at 15:54
• The LHS sum from $k=2$ to $k=n$ agrees with the (expanded) RHS for primes not exceeding $n,$ EXCEPT for RHS terms with denominators greater than $n.$ It is necessary to show that the sum of these "left-over" terms goes to $0$ as $n\to \infty.$ Maybe this is where the deep part is. – DanielWainfleet Aug 23 '16 at 4:05