Visual Harmonic Series Divergence from Bernoulli!

[@ruthvikas]

Nicely explained 🙂

[@MathVisualProofs]

Thank you 🙂

[@Fire_Axus]

Liar

[@ruthvikas]

@@Fire_Axus bruh i study maths in college and this topic is there

[@rblgalichek]

🙂

[@capsey_]

Damn that was clean

[@The9thDoctor]

Nice and elegant. Love it.

[@MathVisualProofs]

Thank you! 😊

[@elidoz9522]

the best proof I've ever seen of this

[@MathVisualProofs]

:) Thanks! Lots of good ones to choose from :)

[@adrianv.v.4445]

Had never seen this proof, beautiful. 👌

[@Ninja20704]

Another way would be to say that the harmonic series is bigger than the area under 1/x from 1 to inf by drawing rectangles, similar to what you did. But since the area, which is really the integral of 1/x from 1 to inf diverges, the harmonic series must also diverge.

[@MathVisualProofs]

Yes. But that proof requires knowledge of integrals (and improper integrals at that ) 😀

[@AlisterCountel]

I usually think of the powers of 2 trick, but this is also a neat proof!

[@MathVisualProofs]

Similar game :)

[@mohammedal-haddad2652]

This is so clever. Thank you very much.

[@diwakarprasad9562]

Nice analogy man...😮

[@muhammadyogaa.9097]

How exciting ! 🤯

[@schrodingersalphacat1862]

Brilliant!

[@Shaeffen_]

Proven to be an infinate sum of 1s even though it seems to be less than 2 😵‍💫

[@stroo_]

1/2 + 1/3 + 1/4 = 13/12, add the one from the beginning, how would it be less than 2?

[@asv3331]

​@@stroo_ at first glance it can look similar to the sum of inverse powers of two. (1+1/2+1/4+1/8+1/16...) which of course converges to two.

[@deleted-something]

That’s interesting, the prime case of a sum that approaches to 0 but still being infinite, otherwise 1/x would have a limited area, and ln(x) couldn’t grow on forever

[@saikatkarmakar6633]

You earned a subscriber

[@prakashkumar1799]

brilliant

[@abdelkarimdebbah9512]

*Math Sharingan unlocked*

[@themathsgeek8528]

Nice!

[@eliasboudjella1141]

The last inequalities seems to be wrong because ln(n)

[@MathVisualProofs]

Why using ln? That’s not part of the justification.

[@eliasboudjella1141]

@@MathVisualProofs . I agree with the proof but i would write the last line in a different way thanks to the partial sums since it is not the same « n » on both sides of the inequality, we sum by packets on the left. It is just the manipulation of infinity which i find weird

[@ThePeterDislikeShow]

I've always learned it as powers of 2.

[@MathVisualProofs]

Yes. Oresme’s proof that you learned is far more prevalent.

[@YOUTUBE_AMERICA]

Love from India you are better than my teacher

[@MochiClips]

I like the 1/2 explanation 1>=1 1/2>=1/2 1/3+1/4 >= 1/4+1/4=1/2 1/5+1/6+1/7+1/8>= 4*1/8 etc... Doubling the number of points and halving the denominator getting an unbounded sum and hence divergent 🗿 I like your argument though it's nice 😎

[@MathVisualProofs]

Yes. That is the classic argument due to Oresme. But this is how Bernoulli made sense of it.

[@kevinbihari]

Sorry man, you lost me Why the graph? What is c? Why 1 over c squared? Hiw do you know what the y coordinate is at x=c squared if c is just a fantasy number? How does that get applied more than 1 times? I really love your style. Would you make a longer video please?

[@MathVisualProofs]

Yes. I was thinking that this could use a longer one so I’ll try to follow up. The value c can be anything. Then you stop at that number squared. So if you use c=26, then you stop at 26 squared. So at the end, you can keep walking along the harmonic series taking the next unused fraction as 1/c and then go to 1/c-squared. Then repeat on the next number serving as the new “c” value.

[@cond.oriano5898]

​@@MathVisualProofs so, this means we can stop at 36, 37etc.?

[@giancarlocafaro6734]

@@cond.oriano5898 ya, exactly. Using the variable c means we can say this about any number, no matter how big. Working with variables makes it easier to manipulate expressions, like when he says (c^2-c)/c^2=1-1/c. You would factor the c from the top expressions, cancel it with one of the c’s from c^2 on the bottom to make (c-1)/c and then distribute the bottom c to make c/c-1/c. That would be hard to conceptualice with numbers.

[@cond.oriano5898]

@@giancarlocafaro6734 we can take any number as variable right? But it should be a typically higher and easier to divide/multiply with number. For our wish, we can take the variable?

[@maxbow-arrow5931]

Great proof! Here's another one: We can replace every element of the harmonic series with a smaller power of two (for example, 1/3 -> 1/4, 1/5 -> 1/8 etc.). Then our sum becomes 1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+8*1/16+...+2^n*1/(2^n+1)+... which is just an sum of infinite halves. This sum is smaller than the harmonic series, so it also diverges.

[@MathVisualProofs]

Yes! This is the classic proof due to Oresme. I find that one to be the most well known so I thought I'd try this one from Bernoulli as a complement :)

[@yangsong6111]

how to come up with the idea to sum till 1/c squared

[@JHnat]

You are going to eventually get to 2. 1+ 1/2… so on and so forth. It will approach if not reach 2

[@MathVisualProofs]

This series gets much larger than 2. It’s slow, but it will be larger than any number you pick.

[@Hassan_MM.]

How does c & c^2 fall on x axis & plz do explain Height of Bottom Rectangle❤

[@MathVisualProofs]

the red (bottom) rectangle has a length of c^2-c on the x-axis. The width of the rectangle is obtained by plugging in c^2 into the function giving 1/c^2. So the area of the red rectangle is the product of these two. Does this help?

[@Hassan_MM.]

@@MathVisualProofs Ya Agreed 👍🏻 but how c^2 & c falls on the same axis, whenever a Degree is added a function seems to fly randomly inshort it no longer remain Linear ❓️

[@10names55]

​@@Hassan_MM. what does it have to do with the plotting of c^2 he is just putting c^2 values for any given c and for the height its just given y=1/x

[@Hassan_MM.]

@@10names55 Don't know why I'm having Difficulty in getting that part

[@wiggles7976]

@@Hassan_MM. The idea is just to start by letting c be any natural number greater than 1 (the edge case of c=1 doesn't really matter). Then, on the x-axis, we plot every number from c to c^2 inclusive. The rest of the argument follows from there. c^2 is just a natural number greater than c, so there's no trouble to plot the function for c, c+1, c+2, ..., c^2 - 2, c^2 - 1, and c^2.

[@feyindecay912]

Sadly I don't understand... does this have to do with music theory?

[@Fire_Axus]

💀

[@wiggles7976]

It has something to do with the fact that a taut string fixed on both ends has a node at 1/1 its length, 1/2 its length, 1/3 its length, 1/4 its length, and so on, for various harmonics. It doesn't actually make much sense to add the wavelengths though; it makes more sense to add the waves themselves through superposition.

[@omnipresentcatgod245]

It's partially related to music theory, however it's very niche

[@feyindecay912]

@@omnipresentcatgod245 I'm more of a composer than a mathematician so I was hoping to learn something interesting about overtones or tuning or something...

[@agytjax]

This proof was originally provided by Nicole Oresme in the 14th century: en.wikipedia.org/wiki/Nicole_Oresme#MathematicsThis short is a visual demo of the same

[@MathVisualProofs]

This is different from Oresme’s proof. This is Bernoulli’s argument (though similar to oresme’s)

[@agytjax]

@@MathVisualProofs - I thought Bernoulli's proof is based on adding the partial sums and arriving at an apparent contradiction a+1=a. Then Bernoulli argues that this is possible only when a is +infinity. Isn't it ? This video is a good visual demo. However, it borrows the concept of constructing a series of partial sums of 2^(n-1) terms of 1/(2^n). Each of the partial series equals to 1/2 and hence an infinite series of 1/2. Here instead of 1/2, you have showed it is >=1. Therefore it looks very much like a Oresme rather than a Bernoulli. Here is one of the several representations of Bernoulli's proof : kzbin.info/www/bejne/g6K5YoCbfcaMj80

[@MathVisualProofs]

@@agytjax I am using this paper scipp.ucsc.edu/~haber/archivesc/physics116A10/harmapa.pdf as a reference (proof 12). I agree that this is similar to oresme’s proof but it is different because of the numbers used and the bounds used (not powers of 2 here).

[@Mike-nv1wz]

you used the associative property to rearrange the terms, partial sums. it is valid if the series is regular ( divergent or convergent) . how do we know the armonic series is regular? i miss this point

[@MathVisualProofs]

This argument can be applied to the sequence of partial sums (where associativity is guaranteed) to show that the sequence of partial sums diverges and thus the series diverges.

[@beath-y5y]

But you cant devide infinite series by parts of sums if it is diverges

[@單邊襪]

As a person who never touched calc Im kind of confused, why cant we just pair up each component of the two series and say that all 1s are bigger than the other half / third / quarter, so the infinite sum of 1 should be bigger?

[@victormagaud2967]

can't we just use the fact that lim(sum(1/k))=integral of ln(x) between 1 and +infinity which diverge ?

[@MathVisualProofs]

That equally seems off. It is bounded below by an appropriate integral of 1/x. Anyway this is an alternate way to prove the fact that does not rely o. Integration.

[@haythemtilouch1191]

I'm curious about how you make theese videos what software u use ?

[@MathVisualProofs]

I use manim for these.

[@erenkayaerenk]

Wow answer is infinity how didn't I guessed?😮

[@bonkerbean6667]

Would 1+2+4+7+11+16+…. Be a harmonic series?

[@MathVisualProofs]

The harmonic series is the sum of reciprocals of all positive integers as shown in this video.

[@bonkerbean6667]

@@MathVisualProofs thank you!

[@wishwellingtonsstuff]

Aaaahhhhhh!

[@yiutungwong315]

Infinity ♾️💍

[@morgangraley1049]

Would this proof still work if you used the left-point? The midpoint? Using a trapezoid instead?

[@MathVisualProofs]

Maybe, but the lower bound might bit work out so clean. Give it a try!

[@vaibhav_topg]

can you do a visiual of ap gp series ? i will subscribe to 😉

[@MathVisualProofs]

I have lots of series videos on the channel. Check kzbin.info/aero/PLZh9gzIvXQUsgw8W5TUVDtF0q4jEJ3iaw

[@ttrss]

real

[@Samy---963]

Why isnt it just 2

[@ZantierTasa]

you're thinking of 1 + 1/2 + 1/4 + 1/8...

[@Samy---963]

@@ZantierTasa ayo wait nah sorry i was wrong

[@Samy---963]

But the technique the guy actually used is lol bruh uncomprehensible

[@ZantierTasa]

@@Samy---963 It's a bit difficult to follow, but if you can understand it step by step, it makes sense. I think this way is a bit easier to understand: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6.... By only REDUCING terms, we can make this: 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + 1/16.... 2 * 1/4 = 1/2 4 * 1/8 = 1/2 8 * 1/16 = 1/2 1 + 1/2 + 1/2 + 1/2... = infinity Because we only reduced terms, the original sequence must be greater or equal to this.

[@yiutungwong315]

​@@ZantierTasa 1 + 1/2 + 1/3 + 1/4 Over 2 So the answer is Infinite Sum 😂🌅

[@saadbenalla3678]

Damn son

[@armandolandomarinelli5087]

What is c

[@MathVisualProofs]

Any number you want it to be… that’s why the argument works because you can keep using the next value for c

[@ujjwalrao6832]

Gah damn that was neat

[@brian8507]

I will pay u 100 dollars if you can give me visualization of 1+2+3+4+5+6+....=-1/12 Visual

[@wumpoooo]

it's false

[@wizparanoid8416]

​@@wumpoooo no it isn't

[@dnuma5852]

@@wizparanoid8416 it is. its in reality whats called a ramanujan summation which assigns values to divergent series. in popular mathematics it has been presented as an actual convergence. it isnt.

[@wizparanoid8416]

@@dnuma5852 isn't there a proof in calculus for Ramanujan paradox?

[@Patrik-bc2ih]

@@wizparanoid8416 It's a false proof. The proof use that 1-1+1-1+...=1/2, And 1-2+3-4+5-...=1/4 Then S=1+2+3+.. 1/4-S=-4-8-12-.... Which is equal to 1/4-S=-4S 1/4=-3S SO, S=-1/12. But this wrong, I added infinite many numbers, and if you sum infinite many number then you cant interchange them. So a+b=/=b+a.

[@rocky_wang]

Good proof but the geometric part seems redundant 😂😂😂

[@MathVisualProofs]

The proof of the inequality is harder without the geometry…

[@ferashamdan4252]

جميل

[@UsernameR8]

The sum is1

[@MathVisualProofs]

This series diverges. So it does not sum to any number.

[@earldominic3467]

what

[@bryanrodriguez2118]

Help! (1÷1)+(1÷2)−(1÷3)+(1÷4)+(1÷5)−(1÷6)−(1÷7)+(1÷8)+(1÷9)+(1÷10)−(1÷11)−(1÷12)−(1÷13)+(1÷14)+(1÷15)+(1÷16)+(1÷17)−(1÷18)−(1÷19)−(1÷20)−(1÷21)+(1÷22)+(1÷23)+(1÷24)+(1÷25)+(1÷26)−(1÷27)−(1÷28)−(1÷29)−(1÷30)−(1÷31)+...

[@znhait]

Proofs like that are just way too random. This is convenient when you already know the behavior of the series, as you can employ any mathematically valid trick. It's not necessarily useful in learning why this series actually converges.

[@MathVisualProofs]

I am not sure of a proof of this that doesn't invoke some (seemingly random) lower bound. The series doesn't converge, it diverges, so you have to find some lower bound. This was Bernoulli's way of finding a lower bound. Oreseme's technique was to look at successive runs of length 2^n and bound those below... a standard modern calculus book bounds the sum below by the improper integral of 1/x from 1 to infinity... they are all similar in some sense.

[@luukderuijter1332]

Soo -1¿

[@loohooi6545]

We've known that 1 + 1/2 + 1/4 + 1/8 + 1/16 + ······ = 1 / 1-0.5 = 2 The harmonic series is clearly greater than the series above,so 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 +……＞2 It is divergent

[@MathVisualProofs]

Why does greater than 2 automatically mean divergent though?

[@yiutungwong315]

1 + 1/2 + 1/4 + 1/8 + 1/16 ... = 1 / (1-0.5) = 2 = π in Riemann Paradox And Sphere Geometry Mathematical Systems Incorporated and That...

[@yiutungwong315]

π^π = 2^2 = 4 = Tau. That's at All 🙂😂

[@verymelone]

my answer to that question was 2 7/60