Why do we assign values to divergent series? For example, the series $1+2+3+4... = -1/12$. I understand the proof for this, but I feel like it uses false math, and I recall reading that you can't do arithmetic/algebra with series that aren't absolutely convergent. Obviously this series diverges, but why do we say it equals -1/12?
Thanks!
The motivation cited in this video involves applications to physics. I shall give an alternative reason.
In G.H. Hardy's book "Divergent Series", a classic on the topic, he gives (on pages 2 and 3) a derivation of the following famous result due to Euler: $$\sum_{i=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ That is a convergent series. The derivation given, due to Euler, uses divergent series! It is true that Euler had other derivations and also true that today we can prove the result using complex analysis without appealing to divergent series. But, just as we can prove results about real-valued functions by appealing to complex-valued functions, so too can we assign values to sequences associated to divergent series in a way that allows manipulations that recover the value of a convergent series. It can pay off to enter a so-called "imaginary world" if the math is made rigorous and there is a way back to whatever is thought of as "real".
Regarding "false math": We might be better off just saying there is an operation (that we shall not call "summation") that assigns values to particular sequences. Then all of these manipulations can be made rigorous in a way that leaves no doubt. It seems that by using the word "sum" in association to the values assigned to divergent series, we create confusion. I leave you with G.H. Hardy's very insightful take on the matter (pages 5-6). The transformations he refers to are divergent series assignments and manipulations of those assignments:
"The results of the formal calculations of [many nice formulae including the one given above] are correct wherever they can be checked: thus all the formulae [...] are correct. It is natural to suppose that the other formulae will prove to be correct, and our transformations justifiable, if they are interpreted appropriately. We should then be able to regard the transformations as shorthand representations of more complex processes justifiable by the ordinary canons of analysis. It is plain that a first step towards such an interpretation must be some definition, or definitions, of "sum" of an infinite series, more widely applicable than the classical definition of Cauchy.
This remark is trivial now: it does not occur to a modern mathematician that a collection of mathematical symbols should have a "meaning" until one has been assigned to it by definition. It was not a triviality even to the greatest mathematicians of the eighteenth century. They had not the habit of definition: it was not natural to them to say, in so many words, "by X we mean Y". There are reservations to be made, to which we shall return [...]; but it is broadly true to say that mathematicians before Cauchy asked not "How shall we define 1 - 1 + 1 - ...?" but "What is 1 - 1 + 1 - ...?" and that this habit of mind led them into unnecessary perplexities and controversies which were often really verbal."
For the zeta function we have the followings:
first $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}, \quad \text{if}\quad {\rm Re} s> 1,$$ and second (due to analytic continuation ) $$\zeta(-1)=-\frac{1}{12}.$$
So, if we "formally put" $s=-1$, we'll get $$\sum_{n=1}^\infty n =1+2+3+4+\dots = -\frac{1}{12}.$$
For how I understand the statement "$\sum_{n\in\mathbb{N}}n=-\frac{1}{12}$" (which is technically false), comes from the fact that we have the Riemann zeta function which is defined for $s\in\mathbb{C}$ with $\Re(s)>1$: $$\zeta(s)=\sum_{n=1}^\infty n^{-s}$$ Now this function admits an (unique, by the identity theorem) analytic continuation, which satisfies $\zeta(-1)=-\frac{1}{12}$.
To answer the more general question, it makes no sense to assign a value to a divergent series, since by using theorems which are valid for convergent series on divergent ones we can obtain basically any value we want (I've seen a proof of this fact, if I find it I will post it). For example we have: $$\sum_{n=0}^\infty (-1)^n=1-\sum_{n=0}^\infty (-1)^n\Rightarrow \sum_{n=0}^\infty (-1)^n=\frac{1}{2}$$ $$\sum_{n=0}^\infty (-1)^n=\sum_{n=0}^\infty [(-1)^{2n}+(-1)^{2n+1}]=0$$ Etc.
See http://en.wikipedia.org/wiki/Divergent_series#Properties_of_summation_methods to see an overview of various methods to "sum" divergent series and desirable properties that summation methods are often demanded to satisfy, like linearity. Under the assumption that a summation method has various desirable properties, you can show for example that
$$\sum_{n=0}^\infty a^n = 1 + \sum_{n=1}^\infty a^n = 1 + a \sum_{n=0}^\infty a^n$$
so that
$$\sum_{n=0}^\infty a^n = 1/(1-a)$$
even if $a > 1$.
