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Motivation (skip if desired): We can show that $$S =\frac{1}{2}+ \frac{1}{2\cdot 3}+ \frac{(2)}{2\cdot (3)\cdot5}+\frac{(2\cdot 4)}{2\cdot(3\cdot5)\cdot 7}+\frac{(2\cdot4\cdot6)}{2\cdot(3\cdot5\cdot7)\cdot11}+...$$etc. converges.

We know

$$\frac{1}{2}\sum_{k=1}\frac{1}{p_k} = \infty$$ and

$$ \sum_{n=1}^{\infty} \frac{1}{n} = \infty$$

and that the sum of factors (in parentheses above)

$$ \frac{2}{3} + \frac{2\cdot 4}{3\cdot 5} + \frac{2\cdot 4\cdot 6}{3\cdot 5\cdot 7}+\frac{2\cdot 4\cdot6\cdot 10}{3\cdot 5\cdot 7\cdot 11}+... = \infty.$$

Question: Let $a(3) = \frac{2}{3}, a(4) = \frac{2\cdot 4}{3\cdot 5},$ etc. Does $$T = \frac{1}{2}+ \frac{1}{2\cdot 3}+\frac{2}{2\cdot3\cdot 5}+\frac{2\cdot 4}{2\cdot3\cdot5\cdot6}+\frac{2\cdot 4\cdot6}{2\cdot3\cdot5\cdot7\cdot7}+~ ...~\text{etc.}$$

$$= \frac{1}{2}+ \frac{1}{2\cdot 3}+\frac{1}{2}\sum_{k=3}\frac{a(k)}{k+2} $$

converge?

Note: $a(n) = \prod_{k=3}^n \frac{p_{k-1}-1}{p_{k-1}}.$

daniel
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  • Is $a(n) = \prod\limits_{k = 3}^n \frac{2k-4}{2k-3}$? – Daniel Fischer Aug 06 '13 at 21:14
  • That should read $\frac{p_{k-1}-1}{p_{k-1}}$, the $p_{k-2}$ in the denominator is a typo, isn't it? – Daniel Fischer Aug 06 '13 at 21:24
  • @DanielFischer: Added a note to the question, thank you. – daniel Aug 06 '13 at 21:36
  • If you know the asymptotic behavior of the main term of $S$ (I do not know), just multiply it by $\log n$ and that will give you your answer. This is because the $n$th prime is roughly $n\log n$. If necessary, use the explicit bounds in this question and its answer: http://math.stackexchange.com/questions/1257/is-there-a-known-mathematical-equation-to-find-the-nth-prime – Lord Soth Aug 06 '13 at 21:39
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    I think the sum diverges because $\log a(n) \sim -\log \log n$, which means the general term is about $\frac{1}{n\log n}$. But it would require a careful estimate to check whether the $\frac{1}{n(\log n)^{1 + \varepsilon_n}}$ you actually get perhaps make it converge. – Daniel Fischer Aug 06 '13 at 21:39
  • @LordSoth: will look at that question. – daniel Aug 06 '13 at 21:42

1 Answers1

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As the commenters have suggested: Mertens proved the asymptotic formula $$ \prod_{p\le x} \bigg( 1-\frac1p \bigg) \sim \frac{e^{-\gamma}}{\log x} $$ (where $\gamma$ is Euler's constant). Therefore $$ \prod_{k=3}^{n} \frac{p_{k-1}-1}{p_{k-1}} = 2 \prod_{p\le p_{n-1}} \bigg( 1-\frac1p \bigg) \sim \frac{2e^{-\gamma}}{\log p_{n-1}} \sim \frac{2e^{-\gamma}}{\log n}. $$ Therefore $$ \frac{a(k)}{k+2} \sim \frac{2e^{-\gamma}}{k \log k}, $$ and so $\sum_k \frac{a(k)}{k+2}$ diverges by the limit comparison test.

Greg Martin
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