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I want to check my attempt at a proof for the divergence of

$$\sum_{n=1}^{\infty} \frac{1}{p_n} \tag{ $\star$ }.$$


We begin with assuming that $(\star)$ converges. If $(\star)$ converges, there is an integer $a$ so that, $$\sum_{j=a+1}^{\infty}\frac{1}{p_j} \lt \frac{1}{b}$$ where $b>1$. Note that given any $b$ there exists an $a$ that satisfies the above inequality. Now, we let $M = p_1\cdot\cdot\cdot p_a$ and consider the number $1+ nM$ for $n = 1,2,\dots$ Any factors of $1+nM$ are $p_i$ for $i \geq a+1,a+2,\dots$ Hence, we can write for each $g \geq 1$:

$$\sum_{n=1}^{g} \frac{1}{1+nM} \leq \sum_{x=1}^{\infty}(\sum_{j=a+1}^{\infty}\frac{1}{p_j})^x$$

But on the right hand side, we have a geometric series: $$\sum_{n=1}^{g} \frac{1}{1+nM} \leq \sum_{x=1}^{\infty}(\frac{1}{b})^x$$ Since the geometric series converges, it means that $\sum_{n=1}^{\infty} \frac{1}{1+nM}$ converges and is bounded above. But that is a contradiction because if we do the integral test on $\sum_{n=1}^{\infty} \frac{1}{1+nM}$, it diverges. Therefore, $(\star)$ diverges.

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    $\begingroup$ Alright! So, from the definition of M, we can say: $\frac{1}{1+nM} = \frac{1}{p_{k+1}p_{k+2}\cdots p_{x}}$, so this term must appear somewhere in the expansion of $(\frac{1}{p_{k+1}}+\frac{1}{p_{k+2}}+\frac{1}{p_{k+3}}+\cdots)^x$ - that's how that inequality came to be. $\endgroup$ – kvmu May 20 '13 at 4:02
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    $\begingroup$ Your statement "Note that as b changes, a changes correspondingly.", though correct, would be better expressed as "given any b, there is an a such that:". This can be valuable in infinite descent proofs, where $a \lt b$,then $a$ satisfies the criterion, so you can apply it again... $\endgroup$ – Ross Millikan May 20 '13 at 4:36
  • $\begingroup$ Oh, good call - I will edit that right now, but other than that is the proof okay? $\endgroup$ – kvmu May 20 '13 at 4:43
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    $\begingroup$ Some nitpicks: Clearly you need $b>1$ and you probably meant $i\geq a+1$ $\endgroup$ – Alex R. May 20 '13 at 5:03
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    $\begingroup$ You might wanna compare your proof to the proof on pg. 18-19 of Tom Apostol's text "Introduction to Analytic Number Theory". I could provide both pages if you like.. $\endgroup$ – user70962 May 20 '13 at 7:51

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