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I've seen the Wikipedia articles on how to sum $1+1+1+1+\cdots=-1/2$ or $1+2+3+4+\cdots=-1/12$.

Is there a theory behind it or is it a random trick? It basically uses analytic continuation of the Riemann zeta?

The article says it may be used in physical applications. So I wonder, what are the conditions that this type of regularization will give me a useful and consistent result?

There are other series which can be summed differently: $1+2+4+8+\dots=-1$. Is it possible to obtain the same result with Riemann regularization? Are different type of regularizations giving the same results (if they exist)? If not, how can I know which one will work for my (physical?) calculation?

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GH Hardy wrote a book "Divergent Series" and there appear also to be some more recent studies available. Others will be able to comment on their direct relevance to your question. Hardy looks at various methods of summing such series in a serious mathematical context. –  Mark Bennet Aug 9 '12 at 9:24
    
Here is the corresponding question for products. –  J. M. Aug 9 '12 at 9:30
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Often regularization can salvage calculations in quantum and theoretical physics involving divergent infinities (certain types of regularization must be chosen). I do not think the reasons precisely why or when it works have been fully fleshed out (see for example my question here or this on Ramanujan summation [not entirely sure of relevance]), but I do expect theory behind this phenomenon to be forthcoming. –  anon Aug 9 '12 at 9:36
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@JM Nitpick: The primes are not (zeta-)regularizable; doesn't mean they aren't susceptible to a different summability method. (Of course the question here is about regularization.) –  anon Aug 9 '12 at 9:42

2 Answers 2

up vote 3 down vote accepted

One of the keys is surely, that any method of summation must be consistent with the results of a conventional sum, if the summation-method is applied to convergent sums/series. I find it fairly obvious (but in case a source in the literature is sought, for instance, Konrad Knopp explained it in his monography on infinite series:) "it must be made clear, what the symbol 1+2+3+4+... really means"; first thing we must do is to find/define an expression for each term in the sum relative to its index. Otherwise 1+1+1+1+... and 1+0+1+0+1+0+... cannot be distinguished of each other. In the geometric series with the base q we have the general term $q^k$ dependend on its index k, in the zeta-series we have the general term as $k^{-s}$ with the fixed exponent s and so on.

So we should first write $s(p)=\sum_{k=1}^\infty a(k,p) $ and define a dependend on k and possibly on another external parameter p to have a description of the general term.

After that we might either find

  • telescoping effects: subsequently following terms cancel (possibly only partially) and the whole expression reduces then to a finite sum
  • continuous intervals for some parameter p, for which $s(p)$ is a convergent series with a closed form expression $g(p)$ dependend on that parameter, where it might be consistent to extend $g(p)$ also for the cases, where the series $s(p)$ would be divergent and so on.
  • $\cdots$

In the latter case it might happen, that we find such an extension and it seems to be a sensical re-expression, but it is found later, that there is also another reformulation, for instance as a sum with a telescoping effect, which reduces then to another value of the (originally divergent) sum. Then that found summation/regularization must be revised - and in general sthis is still a field of open research. There are accepted summation-procedures even for complete classes of infinite series - accessible for instance by Abel,- Cesaro-, Euler-, Borel-, Ramanujan-summation, to name only the classical ones; but there are arbitrarily many series for which we do not have an accepted summation-procedure.
L. Euler's zeta-"regularisation" for instance used that
$$s(p) = \sum_{k=1}^\infty k^p $$ can seemingly be written as $$\sum_{k=1}^\infty (-1)^k k^p +2\cdot \sum_{k=1}^\infty (2k)^p$$ and then $$\sum_{k=1}^\infty (-1)^k k^p +2\cdot 2^p \cdot \sum_{k=1}^\infty k^p $$ and then $$ s(p) = t(p) + 2 \cdot 2^p \cdot s(p)$$ $$ s(p)(1-2 \cdot 2^p) = t(p)$$ $$ s(p) = {t(p) \over (1-2 \cdot 2^p)} $$ where t(p) can approximated for a wider range of the parameter p. But that this works in general depended on, that there is a) a continuous range for the parameter p where this is convergent (and allows the same simplification/reformulation) and b) this range can be continuously extended preserving the meaningfulness of finite values for the $s(p)$ .

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These ridiculous-seeming summations are more a random trick than anything else, but they carry an important lesson. The lesson is that it doesn't always make sense to write $a_0 + a_1 + a_2 + \cdots$ and treat this sum as if it had an actual value. So these ideas are used to teach students that infinite sums don't always exist, or don't always make sense. They're not generally re-visited after first-year analysis class.

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Hardy (and quite a few others) would probably disagree with the first part of your first sentence (which is not needed to support the idea expressed in the rest of your post). –  Did Aug 9 '12 at 9:36

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