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If we say that every number is a product of prime numbers, then we can say that: $$x=\sum_{p\leq x} \lfloor {x\over p} \rfloor-\sum_{p_1<p_2\leq x} \lfloor {x \over p_1p_2} \rfloor + \cdots + 1$$ Now, if we let $x\rightarrow\infty$, we have that: $$x=x(\sum_{p\leq x} {1 \over p} -\sum_{p_1<p_2\leq n} {1 \over p_1p_2}+\cdots)$$ So, is this enough evidence to state the following? $$\sum_{p} {1 \over p} - \sum_{p_1<p_2} {1 \over p_1p_2}+\sum_{p_1<p_2<p_3} {1 \over p_1p_2p_3} -\cdots=1$$

EDIT: Consider now, that the question will be equivalent to saying, (that is, if the absolute value of every term summed is the harmonic series): $$\sum_{n=2\;\omega(n)\not\equiv0\pmod 2}^\infty{1\over n}-\sum_{n=2\;\omega(n)\equiv0\pmod 2}^\infty{1\over n}=1$$ Where $\omega(x)$ is the amount of prime factors of $x$. Which is then equivalent to a weird thing: $$\sum_{n=2}^\infty{-\mu(n)\over n}=1$$ Where $\mu(n)$ is the Möbius function. And as numerical evidence, if you calculate this summation up to $n=10000$, the result will be: $$\sum_{n=2}^{10000} {-\mu(n)\over n}=1.002082699767482251957261311157953789974198663389559562494$$

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  • $\begingroup$ @RoddyMacPhee Can you prove so? $\endgroup$ – Tots Jan 23 '20 at 22:31
  • $\begingroup$ you've summed all inverses of all distinct $n$-almost primes for all $n$ at the very least ... $\endgroup$ – user645636 Jan 23 '20 at 23:09
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    $\begingroup$ Well, $\sum 1/p = \infty$. So, does the third step even make sense? $\endgroup$ – SL_MathGuy Jan 24 '20 at 3:16
  • $\begingroup$ @SL_MathGuy That is my question. $\endgroup$ – Tots Jan 24 '20 at 3:19
  • $\begingroup$ I think the problem must be the very beginning . How did you obtain the equation for $x$ using Sieve of Eratosthenes? $\endgroup$ – SL_MathGuy Jan 24 '20 at 3:45
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Several things are wrong with the step "if we let $x\to\infty$": the asserted equation still has occurrences of $x$ in it, which is not possible if one has taken a limit as $x\to\infty$. If you mean the asserted equation to have limits on both sides, then every one of the sums diverges. In other words, the asserted equation is incorrect for most finite values of $x$ and is not even defined for "infinite $x$".

In short: when we take limits, we have to do so rigorously; we can't use $x\to\infty$ simply as a justification for a mental idea.

Edited to add: it turns out to be true that $$ \sum_{n=1}^\infty \frac{\mu(n)}n = 0, $$ which is equivalent to the "weird thing" in the OP, and is phrased in a way that successfully avoids problems with divergent series. However, this is a pretty deep fact—the typical proof uses the same machinery used to prove the Prime Number Theorem. In particular, it doesn't follow from the sieve-of-Eratosthenes argument given in the OP.

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  • $\begingroup$ I very rarely downvote. I gave this A a +1 and downvoted the Q. $\endgroup$ – DanielWainfleet Jan 24 '20 at 5:31

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