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Starting from the idea that $$\sum_{n=1}^\infty n = -\frac{1}{12}$$ It's fairly natural to ask about the series of odd numbers $$\sum_{n=1}^{\infty} (2n - 1)$$ I worked this out in two different ways, and get two different answers. By my first method $$\sum_{n=1}^{\infty} (2n - 1) + 2\bigg( \sum_{n=1}^\infty n \bigg) = \sum_{n=1}^\infty n$$ $$\therefore ~\sum_{n=1}^{\infty} (2n - 1) = - \sum_{n=1}^\infty n = \frac{1}{12}$$ But then by the second $$\sum_{n=1}^{\infty} (2n - 1) - \sum_{n=1}^\infty n = \sum_{n=1}^\infty n$$ $$\therefore ~\sum_{n=1}^{\infty} (2n - 1) = 2 \sum_{n=1}^\infty n = - \frac{1}{6}$$ Is there any reason to prefer one of these answers over the other? Or is the sum over all odd numbers simply undefined? In which case, was there a way to tell that in advance?

I'm also curious if this extends to other series of a similar form $$\sum_{n=1}^{\infty} (an + b)$$ Are such series undefined whenever $b \neq 0$?


marked as duplicate by user223391, kimchi lover, Claude Leibovici, user99914, Jack D'Aurizio sequences-and-series Nov 7 '17 at 14:40

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    $\begingroup$ I mean the premise that $$\sum_{n=1}^{\infty} n = \frac{-1}{12}$$ is incorrect, so you can't proceed from that. $\endgroup$ – John Lou Nov 7 '17 at 0:37
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    $\begingroup$ @JohnLou there is a notion of summation for which that formula makes sense -- see en.wikipedia.org/wiki/… -- although I don't know enough about it to know what sort of manipulations are valid for that notion of summation (or what the intuitive content of that formula should be) $\endgroup$ – Gregory J. Puleo Nov 7 '17 at 0:42
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    $\begingroup$ Both "the sum over all integers" and "the sum of all odd integers" diverge. As such it's no surprise that when you do things two different ways, you get two answers. If you want a relation analogous to $\sum_{n=1}^\infty n = -\frac{1}{12}$ (in which the equal sign is interpreted with a large amount of liberty) for just the odd numbers, you need to choose a consistent regularization scheme. $\endgroup$ – spaceisdarkgreen Nov 7 '17 at 0:43
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    $\begingroup$ When doing this kind of Ramanujan sum, the values can't be manipulated so easily. For example, $1+(-1)+1+(-1)+\cdots$ does not have the same Ramanujan sum as $1+0+(-1)+1+0+(-1)+\cdots$. You have to be more careful about it than any free-wheeling approach. $\endgroup$ – Thomas Andrews Nov 7 '17 at 0:44
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    $\begingroup$ See Dirichlet Lambda Function. $\endgroup$ – aleden Nov 7 '17 at 0:57

With the usual caveat that $$ \sum_{n=1}^\infty n \ne -\frac{1}{12}$$ we can do a similar zeta function regularization for the sum of odd integers. We start with the fact that $$ \sum_{n = 1}^\infty \frac{1}{(2n-1)^s} =(1-2^{-s})\zeta(s)$$ for $\Re(s) > 1$ and then analytically continue to $s=-1$ to get $$ \sum_{n=1}^\infty(2n+1) "=" (1-2)\zeta(-1) = \frac{1}{12}$$


Zeta function regularization and Ramanujan get the same answer here. As for why your first method gets the "right answer" and the second doesn't, note that the first is argued by the exact same formal steps used to derive $$ \sum_{n=1}^\infty\frac{1}{(2n-1)^s} = (1-2^{-s})\zeta(s)$$ while the second uses both linearity and index shifting which are generally not preserved by the regularization methods.

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    $\begingroup$ You can write it as $\sum_{n=1}^{\infty} n=-\frac1{12}\,(\mathfrak R)$ for a "Ramanujan summation" value. $\endgroup$ – Thomas Andrews Nov 7 '17 at 0:58

If we replace the sum $\sum_{n=1}^\infty n$ by $\sum_{n=1}^\infty n^{-s}$ then (where it converges) we have $$\sum_{n=1}^\infty n^{-s} = \zeta(s)$$ and when $s=-1$ (outside the region where the sum converges) we have $\zeta(-1) = -\frac1{12}$. Since setting $s=-1$ turns the terms of the second sum back into the terms of the first, this is kind of like saying that $1+2+3+\dots = -\frac1{12}$.

Similarly, we can replace your sum by $\sum_{n=1}^\infty (2n-1)^{-s}$. Where it converges, we have $$\sum_{n=1}^\infty (2n-1)^{-s} = (1-2^{-s}) \zeta(s)$$
and if you now substitute $s=-1$ (which is, again, outside the radius of convergence) we get $(1-2) \zeta(-1) = \frac1{12}$. Again, since setting $s=-1$ turns the terms of the second sum back into the terms of the first, this is kind of like saying that $1+3+5+\dots = \frac1{12}$.

But because $(a+b)^{-s}$ is not equal to $a^{-s} + b^{-s}$, once you use this technique to assign values to divergent series, it is no longer valid to say that $$\sum_{n=1}^\infty (a_n + b_n) = \sum_{n=1}^\infty a_n + \sum_{n=1}^\infty b_n.$$ So both of your derivations use technically-invalid operations for this context.


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