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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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    Without really clarifying the proof, the author explained his algorithm and provided extra terms.2012-08-16

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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    Okay I see, thank you. I wasn't suggesting the original series converged. I should have commented rather than answering.2012-08-17
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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.