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In looking at a paper online I came across the following proposition:

$$1 - \frac{1}{9} - \frac{1}{15}- \frac{1}{21}-\cdots = 0$$

After wasting a lot of time, I rewrote it,

$$1 -\left(\frac{1}{9} + \frac{1}{15}+\cdots\right)= 1 - \left(\frac{1}{3}\sum_{k = 2}^\infty\frac{1}{p_k}\right) \rightarrow -\infty. $$

So it seems to me that the r.h.s. diverges. Is this correct?

Edit: in case someone doesn't see Sasha's question and my response here are more terms in the sequence. The denominators are $3 p_k, p_k$ being the $k$th prime. $$1- 1/9 - 1/15 - 1/21 - 1/33 - 1/39 - 1/51-\cdots$$

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    $\begingroup$ There are many sequences with 9,15,21 in it. Why do you think it must be $3 p_k$ one? $\endgroup$
    – Sasha
    Aug 16, 2012 at 20:37
  • $\begingroup$ Without really clarifying the proof, the author explained his algorithm and provided extra terms. $\endgroup$
    – daniel
    Aug 16, 2012 at 20:45

3 Answers 3

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Assuming that the 9, 15, and 21,... stand for $3p_k$, then yes, I would say that this series diverges to $-\infty$.

http://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes

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Infinite series aren't associative in a natural way. Take $1-1+1-1+1-...$ for instance, $(1+(-1))+(1+(-1))+(1+(-1))+... = 0$ but $1+(-1+1)+(-1+1)+...=1$.

Check out the Riemann series theorem it states that you can rearrange a conditionally convergent series so that it sums to any real number or diverges (either to $\infty$ or $-\infty$).

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  • $\begingroup$ You are suggesting the series above may converge as written? $\endgroup$
    – daniel
    Aug 16, 2012 at 20:58
  • $\begingroup$ Everything you say is true, but that is only for infinitely large "regroupings". Here, we can remove the first few terms and note that the rest goes to negative infinity. Then, adding the front on again doesn't change anything. $\endgroup$
    – N. Mao
    Aug 16, 2012 at 20:59
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    $\begingroup$ It's irrelevant. He did not do the thing you described, but merely set apart the first term and multiplied the rest by $3$. Besides, the series has all terms but $1$ negative, so you can do whatever you want with it anyway. $\endgroup$
    – tomasz
    Aug 16, 2012 at 20:59
  • $\begingroup$ Okay I see, thank you. I wasn't suggesting the original series converged. I should have commented rather than answering. $\endgroup$
    – axblount
    Aug 17, 2012 at 12:27
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Your series is a divergent series. To see that, we write the series in the form $$ 1 - \frac{1}{3} \sum_{k} \frac{1}{p_k} \,.$$

Now the series $\sum_{k} \frac{1}{p_k}$ is known as the prime harmonic series which is a divergent series. The study of the prime harmonic series by Euler lead him to discover the zeta function.

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