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<DIV>Actually, that argument doesn't quite work. The convergence of the original sequence is conditional, not absolute, so you can't arbitrarily reorder the terms and draw any conclusions about convergence.</DIV>
<DIV> </DIV>
<DIV>The earlier posted proof (looking at 1/p and 1/(p+1)) is correct; the sequence does diverge. </DIV>
<DIV> </DIV>
<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: Rob Arthan <rda@lemma-one.com><BR>To: hv@crypt.org; santi_spadaro@virgilio.it <santi_spadaro@virgilio.it><BR>Cc: seqfan@ext.jussieu.fr<BR>Sent: Tue, 13 Dec 2005 13:19:12 +0000<BR>Subject: Re: slightly OT: harmonic series revisited...<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_b14ef9f8-20a7-424c-b9e4-8f38b63209b6><PRE><TT>On Tuesday 13 Dec 2005 12:26 pm, <A href="mailto:hv%40crypt.org">hv@crypt.org</A> wrote:
> "<A href="mailto:santi_spadaro%40virgilio.it">santi_spadaro@virgilio.it</A>" <<A href="mailto:santi_spadaro%40virgilio.it">santi_spadaro@virgilio.it</A>> wrote:
> :Anybody knows an answer (and a neat way to show that the answer is
> :true)?
> :
> :"Define a_n = 1/n if n is composite and a_n = -(1/n) if n is
> :prime. Does the series of a_n (sum from n to infinity of a_n) diverges?"
>
> I'm not sure offhand whether P = sum{1/p} diverges, but it doesn't matter.
>...
>
> If P diverges, consider the set {p, 2p, 3p, 4p, 6p}; this avoids collisions
> for all odd primes p, and the contribution to A for these 5 numbers is
> (-1 + 1/2 + 1/3 + 1/4 + 1/6)/p = 1/4p, so A > P/4, and so A again diverges.
>
Nice argument! You only need the second bit because Euler proved that P
diverges. There is some discussion and references at
<A href="http://mathworld.wolfram.com/PrimeSums.html" target=_blank>http://mathworld.wolfram.com/PrimeSums.html</A>.
Regards,
Rob.
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