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Given that $S=\sum_{n=1}^\infty n=-1/12$ (for an explanation see this question or this video from Youtube)

For example if $k=4$:

$(S/4)=1/4+2/4+3/4+1+5/4+6/4+7/4+2+9/4...$

Please edit to improve or if necessary!

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Ellyjant
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1 Answers1

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This has been hashed and rehashed ad infinitum (!) but, obviously,
$$ S=-\frac1{12}\implies\frac14S=-\frac1{48}. $$ And at the same time, since $S$ is $$ S=\sum_{n=1}^\infty\frac1n, $$ then, "obviously", $$ S=\sum_{n\ \text{even}}\frac1n+\sum_{n\ \text{odd}}\frac1n\geqslant\sum_{n\ \text{even}}\frac1n=\sum_{n=1}^\infty\frac1{2n}=\frac12\sum_{n=1}^\infty\frac1n=\frac12S, $$ that is, $$ -\frac1{12}\geqslant\frac12\left(-\frac1{12}\right)=-\frac1{24}, $$ which opens up some fascinating possibilities, such as, sooner or later, $$ -1\geqslant0. $$

Did
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  • I don't think it is valid to say that (S/k)=(S)/k. For example, (S/3)=(1/3+2/3)+1+(4/3+5/3)+2+...=1+1+3+2+...=S+sum_n=1^inf 2n-1 – Ellyjant Mar 30 '14 at 11:51
  • Maybe you did not fully get the take-home message of this answer so let me state it more explicitely: $S=-\frac1{12}$ is a convention, valid in a specific context and only as such. To play with it as if it was a true identity like $\sum\limits_{n\geqslant1}\frac1{2^n}=1$, say, can only lead to chaos. And your question falls squarely into this trap. – Did Mar 30 '14 at 12:20
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    @Did. Great answer, indeed ! May I keep it and reuse ? Cheers. – Claude Leibovici Mar 30 '14 at 12:59
  • @ClaudeLeibovici Thanks. Yes you can (proofs that $-1\geqslant0$ are always free of charge...). – Did Mar 30 '14 at 13:02