Nope🙅 I'll get 0+ ☹️☹️☹️🤭

no, it is 1. except for the first one, every term is approaching to 0, and the number of terms is fixed, because this is the infinite series. And each term is continuous whatever the 's' is. so, the answer is 1. (and when I calculated this by calculator, the result was 1.)

@@user-donghun04 No, every term approaching to 0 may result in any sum

@@user-donghun04 The result is indeed 1, but the method for computing it is incorrect. The correct way to prove it is 1 is by noting that the series converges uniformly on both N and s. Hence the limits can be interchanged.

@@user-donghun04 We know it's 1 but your argument is incomplete. Consider: s(1) = 1 + 1 + 0 + 0 + 0 + 0 +... s(2) = 1 + 0 + 1 + 0 + 0 + 0 +... s(3) = 1 + 0 + 0 + 1 + 0 + 0 +... ... Except for the first one, every term is approaching to 0, and the number of terms is fixed, because this is the infinite series. so, the answer is 1. Except it's actually 2.

Wow, I've never seen that joke before.

Order of seriousness: black -> red -> blue -> green -> purple ?

@@WerewolfLord I’ve never seen purple . . .

@@Camp_RB it's very rare but I think I've seen it a couple times.

@@Camp_RB i think ive seen it in some of his calc vids

@@Gammalol it looks like just a plushie. I remember him using a sphere-shaped mic, but now he might have gotten used to holding something while recording, and the pokeball was the first thing in mind

@@igormatheus8698 It's the mic. You can clearly see the cord

The only animé with an Erdős Number!

@@igormatheus8698 you can see the wire

your feelings are irrational

It can be expanded (carefully) to give the Taylor series expansion of e^x where x=1

Here's a nice explanation of where e comes from, by Eddie Woo: kzfaq.info/get/bejne/ps1oZZqmqbaqoXU.html

But you never are raising it to an infinite power. You’re raising nearly 1 to a massive power. There’s a difference.

@@AngadSingh-bv7vn good luck with the tailor series.

The base is never 1. The base is 1 + 1/n, and 1/n is never 0. Also, the exponent is not infinite. The exponent is n. The exponent diverges to infinity, but is never infinite.

Not -1/12?

@@blackpenredpen It's wrong. So you get docked -1/12 of a mark.That works out in your favour, doesn't it? I think. Ramanujan, where are you when I need you.

@@blackpenredpen well, you are great but not ramanujan

If he lost -1/12 marks that means he got a net +ve

A good point.

I do somewhat hope the explanation will avoid appealing to indeterminate forms and will instead the idea of these limits on its own terms.

At the risk of no one reading, I guess I will lead by example and provide a preliminary explanation of my own. In calculus, there exists the so-called algebraic limit theorem. This is the limit theorem that states that if lim f (x -> p) and lim g (x -> p) both exist and are equal to L and M, respectively, then lim [f + g] (x -> p) = [lim f (x -> p)] + [lim g (x -> p)], and lim [f·g] (x -> p) = [lim f (x -> p)]·[lim g (x -> p)]. This theorem is awesome, because in practice, it is the theorem that allows us to compute the limit of any function at any point symbolically, without needing to use the ε-δ definition of the limit of a function every time. However, the algebraic limit theorem does not hold if lim |f| (x -> p) = ♾ or lim |g| (x -> p) = ♾, because then the limit does not exist for f or g. lim |f| (x -> p) = ♾ is just shorthand for lim 1/|f| (x -> p) = 0. It does not actually mean that lim |f| (x -> p) is literally equal to ♾, it just means that lim |f| (x -> p) does not exist in a special way. Nonetheless, even if the algebraic limit theorem does not hold, it is still possible to symbolically compute several limits. For example, consider lim 0(x)·1/x (x -> 0). 0(x) is the constant function with output 0, so lim 0(x) (x -> 0) = 0, but lim 1/x (x -> 0) does not exist, so it is nonsensical to say that lim 0(x)·1/x (x -> 0) = [lim 0(x) (x -> 0)]·[lim 1/x (x -> 0)]. You cannot multiply by something that does not exist. Yet, lim 0(x)·1/x (x -> 0) can still be computed. Please remember that 0·y = 0 for any real number y. 1/x is a real number, so 0(x)·1/x = 0(x) for any nonzero x. Therefore, lim 0(x)·1/x (x -> 0) = lim 0(x) (x -> 0) = 0. Notice that while the product [lim 0(x) (x -> 0)]·[lim 1/x (x -> 0)] is utter nonsense, lim 0(x)·1/x (x -> 0) = 0. Notwithstanding, if I replace 0(x) by x^1, I get that lim x^1·1/x (x -> 0) = 1, even though lim x^1 (x -> 0) = lim 0(x) (x -> 0) = 0. If lim x^1 (x -> 0) = 0, then evaluating lim x^1·1/x (x -> 0) is like multiplying by 0, and yet we are not getting 0.... except, this is false. Multiplying by x^1 and then letting x -> 0 is not the same as multiplying by lim x^1 (x -> 0), since the algebraic limit theorem does not hold. So multiplying by x^1 and then letting x -> 0 is not the same as multiplying by 0. This is why it makes perfect sense for lim x^1·1/x (x -> 0) to not be equal to 0: you are not actually multiplying by 0 at any point. That being said, teachers have the misleading and inaccurate habit of noting that lim x^1 (x -> 0) = 0 and lim 1/|x| (x -> 0) = ♾, hence lim x^1·1/x (x -> 0) is an indeterminate form of the type 0·♾. The problem with doing this is that it pretends the algebraic limit theorem holds in this situation, even though it is not. One is never actually multiplying by 0, so writing this as 0·♾ is misleading. Similarly, ♾ is not a real number, and one is never actually multiplying by ♾: x^1·1/x is always a real number for nonzero x, and 1/x is always real for every nonzero x. So calling this an indeterminate form of type 0·♾ is inaccurate. What would be accurate is to note that the algebraic limit theorem does not hold, but that the limit may still exist, and that it may be easily computable symbolically. This occurs similarly for lim f^g (x -> p). Remember that if lim ln(f)·g (x -> p) exists, then lim exp[ln(f)·g] (x -> p) exp[lim ln(f)·g (x -> p)]. If lim g (x -> p) = ♾ and lim ln(f) (x -> p) = 1, then we have the sitiation with 1^♾. Many teachers will claim this limit is an indeterminate form of type 1^♾. This is inaccurate, though: ln(f) need not equal 1, and since lim g (x -> p) does not exist, the algebraic limit theorem does not hold, hence cannot say exp[lim ln(f)·g (x -> p)] = exp{[lim ln(f) (x -> p)]·[lim g (x -> p)]} = exp{ln[lim f (x -> p)]·♾} = exp[ln(1)·♾] = 1^♾. Exponentiating g with base f and then letting x -> p is not the same as exponentiating lim g (x -> p) with base lim f (x -> p). Hence you are never exponentiating by base 1. This is clear in the example with lim (1 + 1/n)^n (n -> ♾) = e. Yes, lim (1 + 1/n) (n -> ♾) = 1, but 1 + 1/n is never equal to 1. Similarly, n is never equal to ♾. n is a natural number, so it always finite. A similar analysis works for every indeterminate form, and what it demonstrates is that the indeterminate forms are misleading, which is to say, that whenever a teacher says that a limit is of the form 0/0, you are not actually dividing by 0. You are letting a function f with lim f (x -> p) = 0 be divided by g with lim g (x -> p) = 0, and then let x -> p, which is different than dividing lim f (x -> p) by lim g (x -> p). Only the latter constitutes division by 0. I understand calculus is hard. But it is important that we understand limits and limit theorems on their own terms, and that we stop trying to reduce them to simple "arithmetic" involving the symbol ♾. You cannot do arithmetic with ♾ in the first place. Also, it is incorrect to talk about 0·g when you are actually multiplying by f, and then evaluating the limit, rather than multiplying by 0.

1+dx = 1+ ?

@@POPO-od8jb 1+ could an infinitely small bit (like dx) bigger than 1 if you want, such that it has properties like 1 but doesn't explicitly give indeterminate forms in situations such as 1/ln(x). You can probably define: n+ = inf(n,∞) and n- = sup(-∞,n) if you so feel it needs a definition.

However, we can switch the order of the limits if we note that the terms don't just approach 0 but decrease monotonically to 0.

Exactly what I was thinking. By switching the limits, the "wrong" approach is actually right. I don't think he ecplained it that well.

Ahh u guys probably don't know the rules, he is doing correctly

@@Calculus_is_power You can't always interchange integrals and limits. One counterexample is the sequence of functions f_n(x) = n if 0 oo is identically 0, so the integral of the limit is 0. You can't interchange the integral and the limit in this case.

@@martinepstein9826 ooh yea thanks for clearing my misconcept

I wish that I understood your statement. Can you elaborate on "these"?

ong tho

breaking: wolf becomes chihuahua

@@jasonbroadway8027 The infinite sum implicitly has a limit in it: an infinite sum of f(n) from n=1 to infinity is technically the limit as N->infinity of the sum of f(n) from n=1 to N. So here you have lim_{s->infty}[lim_{N->infty} sum_{n=1}^N 1/n^s] This is not necessarily the same as lim_{N->infty}[lim_{s->infty} sum_{n=1}^N 1/n^s] - you can't just swap the order of limits without further justification.

@@jasonbroadway8027 let me give it a shot. When a sum goes from 1 to infinity, it's shorthand for the limit of that sum from 1 to n as n goes to infinity. So you can take the sum from 1 to 1 and write down the answer. Then write down the sum from 1 to 2, then from 1 to 3, and so on. Youll end up with an infinite sequence. The limit of that sequence is equal to the infinite sum you stated with. This mean that the limit in the video was really equal to this: Limit as s goes to infinity of (limit as m goes to infinity of (the sum from 1 to m of ())). The thing that went wrong in this video is that he swapped the order of the limits. He took each term of the sum and sent s to infinity before he sent m to infinity. That's the wrong order! You need to take the limit in m first and then do the limit in s. This is kinda like the difference between f(g(x)) and g(f(x)) if where f and g are functions. Now, there are some conditions where you can swap the order of the limits. That works when the function you're summing meets a certain "nice" criteria. The expression 1/n^s as a function of n is only "nice" for a few values of s -- certainly not while going to infinity.

🤣🤣🤣

If you use the standard definition of the radical symbol, then no, there cannot be a complex solution to sqrt(y) = -4, because by definition, Re[sqrt(y)] >= 0. sqrt(y) = y - 20 y - sqrt(y) - 20 = 0 [sqrt(y)]^2 - sqrt(y) - 20 = 0 [sqrt(y) + 4]·[sqrt(y) - 5] = 0 sqrt(y) - 5 = 0 sqrt(y) = 5 y = 25.

sqrt(y) = -4 doesn't come from a complex solution, it comes from y = 16 and picking a non-standard square root function (which is admittedly usually conceptually tied to complex numbers)

Why can't y be 25

Woww thats actualy True sqrt25=25-20 can i learn how you did it

You shell not pass

Actually, it is still 1. BPRP is s both right and wrong here.

if 0+ were a real number not equal to 0 then 0+ / 2 would have to be inbetween 0 and 0+

x^2 is not equal to x + x + ••• + x. This is only true for natural numbers x, but the natural numbers correspond to isolated points of the topology of the real numbers, so it is nonsensical to talk about differentiability here.

Yes. Finally someone who knows.

You are indeed worthy of using the username Lelouch

Right, good point. And in this case, the series does converge uniformly on, say, [2,\infty), by the M test, with Mn=1/n^2. So we can switch the two limits

I'm guessing a doctorate

Yeah, when function is continuous 🙄🙄, right 🙄🙄

why you need to talk about religion and god in a math video?

@@timetraveller2818 i want to know does mathematical geniuses believe in god and how

Ιt is 3/2.

Actually, it does converge uniformly. This video is inaccurate. lim ζ(s) (s -> +♾) = 1 is indeed true. You can look this up. The only part about this video that is accurate is the conclusion that the method utilized was invalid (because uniform convergence had not been established), but the explanation behind the conclusion was wrong too.

@@angelmendez-rivera351 I added the word "necessarily" now so if the convergence is truly uniform, I won't contradict this truth.

@@angelmendez-rivera351 It is inaccurate to say that ζ converges uniformly, because it doesn't converge uniformly on its domain, (1,infinity). But it does converge uniformly on [a,infinity) for any a>1, which is enough to justify switching the limits.

@@kyleneilson1457 I see. Now that you commented that, I know how to prove the uniform convergence.

I think you have the right idea. The sum does not collapse to zero, but neither does it blow up to infinity. Instead, it sums up to some finite value like 2/3, 1 or e.

The limit is indeed 1. However, BPRP is still correct in that the method used is invalid, although his explanation for why the method is invalid is incorrect. See my comment to the video for more details.

Why this problem is a little tricky is because there is a tug of war between the terms that are getting smaller and the fact that there is infinitely many of them. Your « explanation » could be reversed the other way around : You could say that eventhough the terms go to zero, they are still bigger than zero, there are infinitely many of them, so the series should not even converge for any s and diverge to infinity. The key for solving these problems is to know how quickly does the terms of the series converge to zero in the neighbourhood of infinity.

@@CrazyPlayerFR That is not an accurate explanation either. There are not "infinitely many " terms. In fact, indeterminate forms have absolutely no relevance in this question. The problem here is that of uniform convergence: we intuitively assume we can just switch the order of the limits and call it a day, but we actually cannot do this in general (though in this particular example, we actually can, making this video a terrible example to illustrate the subject, *sigh*).

@@angelmendez-rivera351 yes i am aware of this, I just posted a comment explaining how we could have done this formally using uniform convergence. I wasn’t giving any « proof » in my comment, just trying to explain why using the same arguements that one could use with a finite amount of terms which all have a limit can not be generalised so easily to an infinite amount of terms. I might have done it wrong though, dont’t know! :).

1^♾ is undefined because ♾ is undefined. ♾ does not denote a mathematical object, and it certainly does not denote an element of an algebraic structure, so talking about the symbol alongside +, ·, and ^ is nonsensical. That being said, for every nonnegative real number L, there exist functions f and g with lim f (x -> p) = 1 and lim 1/|g| (x -> p) = 0, such that lim f^g (x -> p) = L. Hence, the properties lim f (x -> p) = 1 and lim 1/|g| (x -> p) = 0 alone do not allow you to make a conclusion about whether lim f^g (x -> p) exists or not, let alone what nonnegative real number it is equal to. This is why some people say 1^♾ is an indeterminate form. However, this conclusion is inaccurate. The correct conclusion is not that 1^♾ is an indeterminate form, but that the algebraic limit theorem does not hold, meaning that lim f^g (x -> p) does not equal [lim f (x -> p)]^[lim g (x -> p)]. This is because lim g (x -> p) does not exist.

@@angelmendez-rivera351 ty for your explanation 😊

The problem is indeed when interchanging limits. lim lim f(n, m) (n -> ♾) (m -> ♾) is not necessarily the same as lim lim f(n, m) (m -> ♾) (n -> ♾). There is a theorem, called the Moore-Osgood theorem, that indicates the conditions under which the above two are equal.

Of course there are important differences between the zeta series and those Riemann sums, but since the first method doesn't cite any of those differences it can't be correct. It doesn't obviously matter that each Riemann sum is a finite sum, since we can always just rewrite it with an infinite tail of zeros or replace the last term T with T/2 + T/4 + T/8 +... However, no matter how we do this we will never be able to say that for all n the n'th term monotonically decreases from one sum to the next. This turns out to be the key difference.

That’s not possible with elementary function tho.

No wonder you're struggling, it's non-elementary

@@theimmux3034That's the point I was trying to do indefinite non elementary integrals

@@blackpenredpen Sir one youtuber solved Riemann hypothesis this is not a joke really this is link his video kzfaq.info/get/bejne/bt9nmbCeyafHkmw.html

is this joke video? it seems like joke but i cannot tell

The only one to have an Erdős Number!

I do not know what your point was, but there is no such a thing as "Infinity over Infinity". This is utter nonsense.

@@angelmendez-rivera351 its like what the ancient method for the calculation of pi was, you would get closer and closer to the real value but you can't get the exact value, n/n is non definable. So basically you could go up to infinity with it

@@romipog9337 I am not sure of that. lim n/n (n -> ♾) = 1.

@@angelmendez-rivera351 well im not very good at maths but i might add that you should first define n.

@@romipog9337 What? n is a natural number. This much was at least obvious from the video.

1^∞ is undefined lim x→∞ 1^x is defined and =1 Lol

Squeeze the orem

Complex infinity, interesting 🤣🤣

That's what I was wondering as well, not 100% sure

Well, I just want to clarify that that lim ζ(s) (s -> +♾) = 1 is indeed true. What BPRP is disputing is not the answer, but the method for getting to the answer. The issue here is that if you "plug in ♾" and then evaluate the series, you are actually computing lim lim 1/1^s + 1/2^s + ••• + 1/x^s (s -> +♾) (x -> +♾), which is not necessarily equal to lim lim 1/1^s + 1/2^s + ••• + 1/x^s (x -> +♾) (s -> +♾), which is what you _should_ be trying to compute, since lim ζ(s) (s -> +♾) := lim lim 1/1^s + 1/2^s + ••• + 1/x^s (x -> +♾) (s -> +♾). In this case, it just so happens that the two are equal, but switching the order of the limits actually requires proper justification, since in general, changing the order of the limits changes the value.

​@@angelmendez-rivera351 well, I actually commented on the preview of the video as it was showing If you say 1 + 0 + 0 + ... = 1, you'll get zero. When I commented, the video was not released yet :) But now that he premiered the video, it is obvious that he did exactly what I feared: he treats a limit that approaches 0 as zero, which is DIFFERENT from saying 1 + 0 + 0 + ... != 1. So it is basically a clickbait :D The 0 turned out to be NOT the zero that we know and love (at least I love it)!

@@mondherbouazizi4433 I agree with you. In my view, the concept of indeterminate forms is itself clickbait. People say that 0/0 is indeterminate, but this is not the case. 0/0 is undefined. ε/δ, for ε, δ very close to 0, is indeterminate. Similarly, they say 0^0 is indeterminate. It is not. 0^0 = 1. ε^δ, for ε, δ very close to 0, is indeterminate. The same can be said for every so-called "indeterminate form", and really, even the name "indeterminate form" is misleading, since these "forms" represent points in the domain of these functions where a discontinuity exists. This is why I dislike it when maths channels utilize the concept, and I dislike it even more when school teachers bring it up, though admittedly, this is mostly the curriculum's fault. Rather than intuitively illuminate the students, the concept just creates more confusion. I am waiting for the day when I have the time, energy, and resources to start my own KZfaq channel, and dedicate myself to just bust the commonly believed mathematical myths learned within school systems, especially the ones associated with subjects within calculus. Calculus is probably the worst taught mathematics course that I know.

@@angelmendez-rivera351 I couldn't agree more. You summarized exactly all my thoughts regarding this issue! I really really like the way you put it together. And I hope you do create such a channel.. I will definitely subscribe. I actually like the "understand what's going on before diving in" approach. Again, thanks for explaining things in such a nice way. Have a great day!

Bro gamma function proof 0! =1 Gamma(1) = 0! =1

well of course, 0 != 1 anyways

Well of course Since (n-1)! =n!/n And if you put n=1 Then you will get (1-1)!=1!/1 Therefore 0!=1

n! is the product of the n-tuple (1, 2, ..., n). Thus 0! is the product of the 0-tuple (). The product of () is 1. Thus 0! = 1. Q. E. D.

@@numberandfacts6174 Technically, your proof does not show that 0! = 1. Why? Given that identity, Gamma(alpha) = (alpha - 1)!. If alpha = 1, then Gamma(1) = (1 - 1)! = 0! And my question is, why 0! = 1? 😁

an infinite sum is a limit of a partial sum. what you're actually looking at when you look at the riemann zeta limit, is a limit of a limit of a sum. in fact, the main issue of the "wrong" way to do it, is that you're switching the order of those two limits (which can be done sometimes, but not always). basically, the numbers of terms is dependant on n in both examples.

@@tabeh- how will it be dependant on n(or s for first example)? the infinite sum ,which as you say can be defined as the limit of a partial sum, here let’s say the number of terms is b , why should b and n be dependent on each other?

@@manik1477 basically [lim(s->inf) Σ (from 1 to inf) 1/n^s] can be written as let's say [lim(s->inf) lim (N->inf) Σ (from 1 to N) 1/n^s]. hard to show this through text, but this is as clear as it gets. So essentially you're taking a sum of N terms, then the limit of N -> inf and THEN you take the limit of s -> inf. There is a limit inside the sigma notation. The partial sum is not really dependant on n, but some other letter that I called N. When he does the "wrong" way of calculating it, instead of [lim(s->inf) lim (N->inf) Σ (from 1 to N) 1/n^s] he calculates [lim(N->inf) lim (s->inf) Σ (from 1 to N) 1/n^s] that is, by using the s limit first he gets the sum 1+0+0+0 of N-1 zeroes, which he then continues to infinity by using the N limit.

@@tabeh- Oh okay, I get the picture now,Thank you so much for explaining this!

Wrong. It is indeterminate when the 0 is not a 0, but rather a limit that approaches 0. The definition of 0 says explicitly that it is the neutral element with respect to addition in IN or IR, thus no matter how many times you add it, the result won't change. In other words, if your "zero" is the limit as x approaches infinity of 1/x, then yeah, that's indeterminate. However, if it is the real 0 that cancels anything you multiply it by, and adds literally nothing (not even a tiny infinitesimal fraction), then your logic is wrong unfortunately. There is something a bit deeper here. Not sure what trick Mr. Blackpenredpen hides, but it's not your simple transformation of addition into multiplication.

@@mondherbouazizi4433 Thank you for clarifying that. Appealing to indeterminate forms is so misleading here.

1/∞=0 but sometimes 0+

This is not a proof and is just informal intuition but may answer your question: Say the notation for 0.9 with n repeating decimals is 0.9_n. So 0.9_1 is 9/10, 0.9_2 is 99/100, and so on. We will say that 1 minus this value is 10^-n which you can confirm for yourself by checking 0.9 + 0.1 = 1, 0.99 + 0.01 = 1, etc. So we have that 1 - 0.9_n = 10^-n, or 1 = 0.9_n + 10^-n, and we will take the limit as n approaches infinity on both sides. Lim n-->inf 1 is still exactly 1. Lim n-->inf (0.9_n + 10^-n) = 0.999 repeating + 0. You may feel inclined to argue that Lim n-->inf 10^-n is actually 0^+ as discussed in the video, but since it is not a term in an infinite sum, it equals exactly 0. So, 0.999... equals exactly 1.