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When I see the answer to the problem. How to check the convergence of the series whose elements are taken from the set $A$?

One question raised in my mind, Can I prove $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges using the following technique?

By the Prime decomposition theorem, $n=2^{j_1}3^{j_2}...$

$$\sum_{n=1}^{\infty}\frac{1}{n}=\sum_{j_1=1,j_2=0,...}^{\infty}\frac{1}{2^{j_1}3^{j_2}...}=\sum_{j_1=0}^{\infty}\frac{1}{2^{j_1}}\sum_{j_2=0}^{\infty}\frac{1}{3^{j_2}}...=\frac{2}{1}.\frac{3}{2}.\frac{5}{4}...$$

Can I judge using the above technique? Where did I go wrong?. Please help me.

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    The idea is fine as exposed here (actually this is a way of proving infinitude of primes !). – Raymond Manzoni Nov 19 '17 at 14:19
  • @RaymondManzoni I really wonder, How could I think like this.!!!. thank you for the reference. –  Nov 19 '17 at 14:24
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    This might also be of interest, same idea but in my opinion a little further explained. – kingW3 Nov 19 '17 at 14:56
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    You still have to prove that the product $\prod \left(1+\frac{1}{p}\right)$ took over primes is divergent, which is quite the same as proving that $\sum \frac{1}{p}$ is divergent. True, but non-trivial at all. – Jack D'Aurizio Nov 19 '17 at 20:55
  • Theorem: If $a_n\geq 0$ then $\sum_na_n$ converges iff $\prod_n(1+a_n)$ converges. See the above comment by Jack D"Aurizio – DanielWainfleet Nov 20 '17 at 01:29

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An interesting remark is that due to Euler's product

$$ \forall s>1,\qquad \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s}=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^s}\right)^{-1} $$

we are allowed to state:

Since the harmonic series is divergent, there are infinite primes.

Of course you may also state

Since there are infinite primes and their sequence does not grow too fast, the harmonic series is divergent

and prove through elementary means that $\pi(n)\geq \frac{n\log 2}{2\log n}$, but this is nuking mosquitoes.
The previous claim can also be stated as

Since $\sum_{p\in\mathcal{P}}\frac{1}{p}$ is divergent, $\sum_{n\geq 1}\frac{1}{n}$ is divergent as well.

Jack D'Aurizio
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