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If $1+2+3+4... = -1/12$

then, $(1+2+3+4...)*1/2$ should equal $-1/24$

But I find this strange since the second infinite is larger than the first because $1/2+2/2+3/2+4/2\dots$ contains all integers of the first group $(2/2,4/2,6/2,\dots)$ plus all the other fractions. But the sum of all its components is smaller.

I am seeing this wrong?

Simply Beautiful Art
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Cruclax
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  • Please use $\LaTeX$ on this site, it makes it easier to read the problem. – Simply Beautiful Art Mar 23 '16 at 23:57
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    You are using consciously the fact that divergent series can be manipulated by multiplication,addition and the like. Of course you are going to get a result like this if you proceed likewise. You just cannot say that $1+2+3+4=-1/12$, because that result has been calculated using the mathematics of divergent series, which isn't accepted in general, but is talked about like a fairytale, as if it becoming true would be the best thing to happen in the world. As far as that theory goes, I'm happy we have counterexamples like the one you have given. – Sarvesh Ravichandran Iyer Mar 23 '16 at 23:57
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    What you are describing is the Ramanujan summation convention, and the values assigned to divergent series in this way do not act in the way we would like them to. In other words, the number $-1/12$ is associated to the series $1+2+3+...$, but it is wrong to assume that the series equals $-1/12$ in the usual sense. So we cannot assume that $1/2+2/2+...=-1/24$, and even if it did, it is not a comment on the "size of the infinity" of each series' divergence. – Plutoro Mar 24 '16 at 00:01

1 Answers1

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You can't really perform math on a divergent series and expect things to work logically.

For example, there are certain properties that a series might lose, most of these relating to the way we can manipulate the series and still get logical results.

Since the result of $-1/12$ was found in a way that made manipulating the series difficult, we cannot expect our result to be manipulatable in the common sense.

Simply Beautiful Art
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