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I just found out that the Riemann Series Theorem lets us do the following: $$\sum_{i=1}^\infty{i}=-\frac{1}{12}$$But it also says (at least according to the wikipedia page on the subject) that a conditionally convergent sum can be manipulated to make it look like it converges into any real number. My question is then: Is there a general algorithm for manipulating this series into an arbitrary number?

My knowledge about series and number theory is pretty limited so if I'm in over my head or if the answer is just too complicated I'd appreciate some tips on what to read up on!

Thanks!

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6  
Does the Riemann Series Theorem let you do that? The theorem requires a conditionally convergent sequence. I don't see how $\sum_{i=1}^\infty$ is conditionally convergent. – David K Mar 28 at 17:49
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The Riemann Rearrangement theorem requires that your sum $\sum_{i} a_i$ converges but $\sum_i |a_i|$ diverges, which generally means that $a_i$ take on positive and negative values (which makes summing to a negative number more reasonable). This begs the question however, of whether there exists a rearrangement theorem for Ramanujan summation. – Alex R. Mar 28 at 17:49
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The sum in question is not conditionally convergent, under the definition employed by the Riemann Series Theorem...therefore that theorem does not apply. The summation you cite employs an alternate interpretation of the sum. – lulu Mar 28 at 17:49
    
Riemann Theorem needs a series whose partial sums converge to begin with. – Alexandre C. Mar 28 at 19:43
    
up vote 8 down vote accepted

The theorem itself is proven by giving the algorithm. You can find a proof on Wikipedia: https://en.wikipedia.org/wiki/Riemann_series_theorem

However, the sum of the positive integers doesn't converge, no matter what order you put them in. The -1/12 result comes from a broader notion of summation than convergence, and is not connected to the Riemann rearrangement theorem.

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I tried that but i couldn't get it to work. I assumed that the series of positive integers was the absolute value version of this series. Does this broader notion of convergence you speak of have a name i can look up? – Zluudg Mar 28 at 17:54
    
The sequence 1 - 2 + 3 - 4 + ... isn't conditionally convergent either. There are various methods for assigning "sums" to divergent series, and I'm not an expert on how they interrelate, but you could look up Ramanujan summation and zeta regularization. – Daniel McLaury Mar 28 at 18:07
    
Will do! Thanks for the tip! – Zluudg Mar 28 at 18:10

There is such an algorithm, but first a quick note: $\sum_{i=1}^\infty i$ is not a conditionally convergent sequence. It is not convergent at all. It is equal to $-1/12$ in the sense of Ramanujan convergence, but that is not applicable to the Riemann series theorem.

Suppose that the series $$\sum_{i=1}^\infty a_n$$ of real numbers is conditionally convergent. Then there must be an infinite subsequence $\{a_{n_k}\}$ of positive terms of $\{a_n\}$ and an infinite subsequence $\{a_{m_k}\}$ of negative terms of $\{a_n\}$. The sequence $\{a_{n_k}\}$ must ahve a largest element, and $\{a_{m_k}\}$ must have a smallest element. Let's say you want to rearrange the series to add to some number $r$. Suppose that $r>0$. Start adding the terms of $\{a_{n_k}\}$ starting from the largest on down until the sum is bigger than $r$. Then add terms of $\{a_{m_k}\}$ starting from the smallest on up until the sum is smaller than $r$. Then add some more terms of $\{a_{n_k}\}$ until the sum is bigger again, and continue the process forever. Since the series is conditionally convergent, both series $$\sum_{k=1}^\infty a_{n_k}$$ and $$\sum_{k=1}^\infty a_{m_k}$$ are infinite, so you can always complete each step. In the limit, your sum will be $r$. If you want to add to some $r<0$, do the same thing, but start with the negative terms instead.

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Yeah, i recognise this from the wiki-page on the Riemann series theorem, but i understand now why the theorem is not applicable to my situation; there is no subsequence of negative terms and no greatest number. Maybe i should read up on Ramanujan convergence instead! Thanks! – Zluudg Mar 28 at 18:09

The Riemann series theorem does not allow you to make the claim above because $ \sum_{n=1}^{\infty} n$ is not a conditionally convergent series.

Instead, an amazing but customary abuse of notation is used to write down this "identity". There is a function called the Riemann Zeta Function which is defined for every complex number except $s=1$. If $s$ has real part greater than $1$, the value of the Riemann zeta function equals $$ \zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}. $$ The Riemann zeta function also has $$ \zeta(-1)=\frac{-1}{12}. $$ Filling in $s=-1$, we get $$ \zeta(-1)``=''\sum_{n=1}^\infty n, $$ but this should not be taken too literally because the sum only converges when $s$ has real part greater than $1$ and does not hold if $s=-1$.

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