Can someone explain the meme that you're stealing 1/12 from me in this sum?

Okay, good job.

There is a theorem by Euler expressing integrals as sums (resp. sums as integrals). He came up with it to compute hard integrals (resp. hard sums). It is

[math] \int_a^b f(n)\,{\mathrm d}n = \sum_{n=a}^{b-1} f(n)+\left(\lim_{x\to b}-\lim_{x\to a}\right)\left(\dfrac{1}{2}-\dfrac{1}{12}\dfrac{d}{dx}+\dots\right)f(x) [/math]

If, for example, [math]f(n)[/math] is of order [math]n^{k-1}[/math], then the left hand side (integral mapping [math]n^{k-1}[/math] to [math]n^k[/math]) is something of order [math]b^k[/math]. The right hand side are still only of order [math]b^{k-1}[/math], but the price is that it's a lot of terms. The coefficients come from the even Bernoulli numbers.

So for example, with

[math]f(n)=n^{k-1}[/math]
[math]k=3[/math]
[math]a=2[/math]
[math]b=4[/math]

you get the identity

[math] \frac{1}{3}(4^3-2^3) = (2^2+3^2)+\frac{1}{2}(4^2-2^2)-\frac{1}{12}2(4^1-2^1) [/math]

You couldn't make this shit up.

For k=2, a=0 you can easily visualize the result: The integral [math]\int_0^b n\,{\mathrm d}n[/math] is the surface under [math]f(n)=n[/math], i.e. the triangle area [math]\frac{b^2}{2}[/math]. The sum of [math] 0+1+2+ ... +(b-2)+(b-1) [/math] corresponds to the surface under a staircase. Hence for ever step, the sum misses [math]\frac{1}{2}[/math] for each step. You get [math]\int_0^b n=\sum_{n=0}^{b-1} n-\frac{b}{2}[/math], which you might write as [math]\sum_{n=0}^{b-1} n=\frac{b(b-1)}{2}[/math].
Back to the formula. It would tell us directly:
[math]\sum_{n=0}^{b-1} n = (\frac{b^2}{2}-\frac{b}{2}+\frac{1}{12})-(0-0+\frac{1}{12})=\dfrac{b(b-1)}{2}[/math].

A summation technique is a procedure taking a sequence of numbers to a single number. E.g. the sequence 1/2, 1/4, 1/8,... given by [math]f(n)=1/2^n[/math] may be mapped to 1 (the result of the classic sum in analysis).
Ramanujan summation is roughly the Euler formula, except the icky
[math] \lim_{x\to b} [/math]
is dropped. But then f(n):=n, with f'(n)=1, pics up a Bernoulli term -1/12.

Just grab a calculator and start adding up the numbers:
1+2+3+4+...
Eventually it'll display 0.08333333... which is 1/12.

This was discovered in the '70s, and was originally caused by a register overflow bug in old HP-35 calculators, but it wasn't discovered until the '90s. However, at that time too many mathematical theorems have been based on this result, so it would have been too expensive to correct the mistake.
Since then it became sort of an inside joke among mathematicians, and calculator manufacturers pay homage to the much revered HP-35 by handling the special case of 1+2+3+4+... = 1/12 on nearly every calculator today.

>1/12
I meant -1/12

Hah, this.

>not [math]\displaystyle\sum^\infty_{i=1}i \stackrel{\scriptstyle \Re}{=} - \frac{1}{12} [/math]
step it up senpai

>So all ramanujan sums are a relation to the original sum (1-1+1-1....)?
No.

This is advanced trolling lol.

youtu.be/jcKRGpMiVTw
youtu.be/w-I6XTVZXww

All you need to know

Pretty trivial desu senpai

What does the "R" symbol mean?

Read OPs question again and have an educacted guess.

>ramadan sums
>negative numbers
That shit don't add up.

>Your post cannot be considered a proof
Of course he hasn't proven anything, the end result is completely untrue. Each step make intuitive sense (if you understand basic algebra), so one might hope that the result has some significance, which it does.
Using this sum to assign finite values to divergent quantities in a systematic fashion is called zeta function regularisation, and is used in number theory and theoretical physics.
The fact that defining the sum of all natural numbers as -1/12 can be used in physics in a way that agrees with experiment shows that there must be something important going on here, regardless of your fee-fees.

Via Ramanujan summation

Don't be a shitbag, it's easy to have very disconnected knowledge of Ramanujan summation if you've spent hundreds of hours reading things you don't have the prerequisite knowledge for. I knew about analytic continuation but that notation flew over my head the first time I saw it.

As someone has already said, it's the Ramanujan summation: en.wikipedia.org/wiki/Ramanujan_summation

It's easy to confuse yourself with this shit but it's quite simple.

All that the ramanujan summation stuff, cutoff and zeta regularization does, is look at the smoothed curve at x = 0.
What sums usually do is look at the value as x->inf.

It's just a unique value you can assign to a sum, really they have many such values.

en.wikipedia.org/wiki/1_+_2_+_3_+_4_+_⋯#/media/File:Sum1234Summary.svg

So it's another method of adding up numbers that in this special case produces that value...

Hello highschooler, how are you enjoying numberphile?

First there's the idea of writing a sum as an integral plus a correction term. Assume a(k) has a taylor series around k=0.
[math]\displaystyle \sum_{k=0}^\infty a(k) = \int_0^\infty a(t) dt + \sum_{n=1}^\infty a^{(n-1)}(0) \frac{\mathrm{B}(n)}{n!}[/math]
Notably if the sum of a(k) is easy to integrate then the series can be replaced with
[math]\displaystyle \sum_{k=0}^\infty a(k) = \int_0^\infty a(t) dt + \int_0^1 \sum_{k=0}^{t-1} a(k) dt[/math]

Now the problem with Ramanujan summation for divergent sums is that it proposes dropping the integral outright instead of regarding it as something that also needs to be carefully extended outside of it's normal domain of convergence.
In principle doing this isn't "wrong", it is what it is, so the question is whether it's a meaningful way to sum a divergent series.

A simple test is to see if it agrees with the regular sum where it actually converges. We can see that's not the case by looking at the geometric series. The Ramanujan sum of [math]a^k[/math] is [math]\frac{1}{\log(a)} + \frac{1}{1-a}[/math] while we would expect to get just [math]\frac{1}{1-a}[/math].

But Ramanujan summation does seem to work for the zeta series. That's because [math]\int_0^\infty t^s[/math] actually happens to be 0 when extended to positive s, so it doesn't matter if we leave it off.
I've looked into this for s=2 and came up with the following (disgusting) regularized integral
[math]\displaystyle \lim_{x \to 0} \int_0^\infty \left( t(t+1) (1 - x/(x+1) (t+2)/3)/(x+1)^{t+2} - t (1 - x/(x+1) (t+1)/2)/(x+1)^{t+1} \right) dt[/math]
If you type it into Wolfram Alpha you'll see that it actually converges to 0, and if you were to set all the x's to 0 you end up with an integral just over t^2.

If we go back to the geometric series example
[math]\displaystyle \lim_{x \to 0} \int_0^\infty e^{-t x} a^t dt = \lim_{x \to 0} \frac{1}{x-\log(a)} = -\frac{1}{\log(a)}[/math]
Which explains where the extra log was coming from in the Ramanujan summation.