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The engineers were right all along! Taking the average does give the true answer.

I like the fact that most different approaches lead us to the same ridiculous answer.

I think maths gets weird at infinity. As an engineer, I'm happy to accept the idea of limits, calculus and converging sums. But I give up on trying to justify values for divergent sums.

Mathologer did a good lesson on this one a few years ago. It's all fun and games but you can run into contradictions pretty easily so it's better to be very careful about it. Most important observation is that convergence of analytic continuation does not imply convergence of the original expression. That process may only work one way.

Ahhh, Cesaro sums, the annoyance of my Reals class

"Any math which uses letters is witchcraft."  --George Washington

Mathologer and 3blue1brown have also both done videos on “sums” of divergent series that are pretty great. Basically you’re essentially extending the usual definitions of “sums” to assign a number that makes sense in a variety of contexts to these divergent sums. The famous 1+2+3+... = -1/12 for example makes no sense normally but, if you add a formal way to assign a number to that divergent sum, then that -1/12 makes sense and tells you something useful about the series itself. Likewise saying that 1-1+1-1+... = 1/2 doesn’t make sense in the usual settings, but when you extend the concept of summation so that you can assign a value to that series then 1/2 pops out as a sensible value.

It’s a bit like how square roots of negative numbers make no sense in the Reals, but if you extend the number system to define and assign complex numbers to square roots of negative numbers then the resulting numbers make sense in that context. Or when you analytically extend the definition of factorials to be the gamma function, you have a new function that still gives you the normal values for factorials but also gives you useful values for other numbers.

Mathologer - https://youtu.be/jcKRGpMiVTw

3blue1brown - https://www.youtube.com/watch?v=XFDM1ip5HdU

I have another example where this sum shows up, which  I find pretty interesting. The integral of sin(x)*cos(x) is 1/2sin^2(x). But if you perform integration by parts you find that the integral of sin(x)*cos(x) is equal to sin^2(x) - that integral. So if you continue the integration by parts and factor out sin^2(x) you get the sum (-1)^n times sin^2(x) and therefore the relation that this sum is equal to 1/2.
Edit: I ignored constants while integrating since this makes no difference in my argument.

"I was about to give up... but!!!"

The answer 1/2 makes sense to me from a logical standpoint since for the finite series the sum is dependent on the mod 2 result of the number of terms. Given that infinity is neither (or both, depending on your frame of reference) odd nor even, and the convergence point is by definition a single value, the only way that it can represent both of them simultaneously is by their midpoint.
By extension, for any series with a repeating pattern of sums the convergence point should be the sum of the vales in the repeating set over the number of values.

Guido Grandi is like ariana grandi of maths

tis is what i will answer when they ask me to find Schrödinger's cat in maths

General tests of divergance is always telling the truth. The one thing happens when you use these kind of formulas and get 1/2 means that "if" this series did converge, it wouldn't be anything other than 1/2.

Here is an idea to consider: the sequence f : N —> R with f(n) = (–1)^n has a sequence of partial sums equal to (s[f])(n) = [1 + (–1)^n]/2 = 1/2 + f(n)/2. So lim s[f](n) (n —> ♾) exists if and only if lim f (n —> ♾) exists. Of course, we know that lim f (n —> ♾) does not exist, so lim s[f] (n —> ♾) does not exist either. However, if lim f (n —> ♾) = 0 were true, then lim s[f] (n —> ♾) = 1/2 would be true as well.

This motivates the concept of regularization. Rather than evaluating lim s[f] (n —> ♾), it may be interesting to look at some other linear operator of sequences instead. This linear operator, call it L, should coincide with lim f (n —> ♾) whenever it exists, but it should have the added property that L(f) = 0 if f(n) = A·sin(B·n + C). Ideally, it should also be a continuous operator. Then L(s[f]) would be the quantity of interest. Just a fun experiment. I believe at least one such operator L does exist.

I really like that he's probably the only YouTuber who asks his viewers to dislike the video if they don't like it.

I am a beginner at analytic continuation. To me, it doesn't really seem like the analytic continuation is even the same function we started with (that, or it's more a piecewise function on a different domain). The Taylor Series we used to analytically continue 1/(1+x)  does NOT equal 1-1+1-... when we substitute x=1, yet it does seem like there's some sort of link between the two. I would really appreciate if someone could explain this a little bit better and give some insight into the validity of this method.

Regardless, I hope you enjoyed the video! If you didn't enjoy it, make sure to give it a dislike! :)

It is interesting that if instead you look at -1 + 1 - 1 + 1 -1 + ... then you end up at a sum of -1/2  (using the same reasoning).  So although it is an infinite series, the sum result depends on which is your starting value (either +1 or -1)

In school, I once did a question in which there was limit to infinite and after few minutes my small brother saw my question Nd told that the answer was the average of the original range of the function. And we tried to apply it on few more questions and it worked. I don't remember the question, but after that we were kinda curious why that happened.
This seems like an explanation to those problems 👀

We are just calling it summation but in fact doing something different. With odd number of terms we get 0, with even number of terms we get 1. Hence limit as n approaches infinity of this series doesn't exist. By changing and modifying definitions to suit our needs we can make it possible for the series summation to equal 1/2. But then we aren't even summing the damn sequence. We have changed the definition of summation.