Simple Analogy to explain $\sum_{n=1}^\infty n = -1/12$ I'm looking for a simplified analogy to explain why the following formula does not actually mean what it seems to mean:
$\sum_{n=1}^\infty n = -\frac{1}{12}$
I get this question all the time from high school students, probably because it was popularized by Numberphile on Youtube. (See video here and see associated post here)
Since the concept of power series expansion is tough for a lot of high school students to get (let alone the zeta function or analytic continuation), I'd like to give them some accessible way to understand how that equal sign isn't really doing what we normally think it is.
I am partially indebted to the answer of user3002473 here.
Please let me know if my understanding of the problem is correct and if my analogy makes sense and is apt:


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*My understanding of the problem:


The zeta function exists everywhere in the complex plane, but its power series representation is valid only on a portion of the plane.
The equation $\sum_{n=1}^\infty n = -\frac{1}{12}$ is therefore the result of equating the "analytic continuation of the function" at a point outside the power series representation domain with the numerical value of the power series computed at that point.
In other words, each side of the equal sign is a valid representation of the zeta function, but neither representation is actually trying to communicate the numerical value of the function at that point.  


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*My analogy:


Consider a non-vertical line in the plane with slope, $2$, and y-intercept, $3$.  We can describe this line as a function between two sets of numbers, $f:X\rightarrow Y$
Clearly, we can see that the graph is one visual representation of the function, but we can also represent it as an equation in slope-intercept form: $f(x) = 2x+3$
Now, suppose I wanted to represent this line as a function with $x$ in the denominator.  I could do that like this:
$g(x) = \frac{2x^2+3x}{x}$
This is clearly a valid representation of my linear function, but it breaks down at the point $x=0$, like so:
$g(0) = \frac{2\times 0^2+3\times 0}{0} = \frac{0}{0}$
As we all know, $\frac{0}{0}$ is "undefined".
Key point:  The original function exists at $x=0$, but the specific representation $g(x)$ fails at that point.
Now, I could say, "We all know that $\frac{0}{0}$ is undefined.  But for this function, we have enough information to 'define' it."
That is, we know that in the original function, the input $x=0$ is associated uniquely with the output $3$.
So, since $g(x)$ is a representation of our function, we could (strongly waving our hands) say that the output $g(0) = 3$
But then, we've just "proved":  $3 = \frac{0}{0}$
No, not really.  All we did was take equate two different representations, each valid over different domains, and equate them at a point they don't have in common.
Is this a reasonably accurate analogy to the trouble with $\sum_{n=1}^\infty n = -\frac{1}{12}$ ?
 A: You can do all kinds of things like that, by plugging in numbers to a series outside its radius of convergence. You can take the Taylor series for 1/(1-x) & plug -1 in & 'prove' that $$\sum_{k=0}^\infty (-1)^k=1/2 $$ (!) It all just emphasises the importance of keeping track of convergence, really.
The particular example you cite is a result of the analytic continuation of the Riemann zeta function beyond the domain on which its representation as $$\sum_{k=0}^\infty {1\over{k^z}}$$ is valid. When one exhibits a pathological case such as the one you cite at the beginning of the question, one is just taking a result produced by a new recipe & projecting it back by sheer force onto the old, which no-one ever did say (at least no-one who understands the matter aright) was valid in the domain inwhich the new one has produced its result. And yet there is a continuity between the new & old recipes: they splice together seamlessly. 
Convergence & analytic continuation are indeed profound matters ... and yet ultimately all you are doing is counting!
