I've had this question for a while. Thank you for this simple but thorough proof. Now I can apply the same methodology to understand why alternating harmonic series converge.
*Important question…* In this video, all the denominators in the series are put to the power of 1, and the series diverges. HOWEVER, in the Basel equation, all the denominators are put to the power of 2, and the series converges. *question:* As the denominator powers transition from 2 down to 1, at what threshold power value will the series likewise transition from converging to diverging?
Why the divergent harmonic series cannot proof Zeno Paradox of movement? Why is proved with the series of 1/2^n and not with that of 1/n if both sequences tends to 0 ? Ok series 1/2^n is convergent and our harmonic divergent, but if Zeno asked for 1/2, 1/3 ,1/4, 1/5 ,...,1/n to the destination, this divergent is one case Against the other convergent used to proof of movement. Is not a "math cheat" choose one sequence convenient to the proof without explain why others are not valid AGAINST that choosen? If the answer is only :"divergent is infinite sum" this gives reason to Zeno, as he have one valid infinite sum against that convergent proof. Who wins? I think have math answer to this, but post my initial fair though. PS: i see in the convergent as in each step summing 1/2^n to the acquired also remains 1/2^n to destination, as in the harmonic summing 1/n to the acquired remains (1-1/n) to destination.
I'll assume you're talking about Zeno's most popular paradox, _Achilles and the tortoise_ You can cover a distance of 1 by using _finitely many_ terms of the harmonic series (even excluding 1), in this case 1/2 + 1/3 + 1/4 > 1. Zeno considered _infinitely many_ steps of 1/2 + 1/4 + 1/8 + ... length to cover the distance of 1. He argued it'd take an infinite amount of time to complete this process. The paradox is resolved when we observe the time it takes to complete a step also decreases geometrically: 1/2 + 1/4 + 1/8 + ... units of time (assuming constant velocity, of course); therefore, a distance of 1 will be covered in 1 unit of time.
@@urthoperator3126 Thanks, you are right. What I said is about the real line being the same, the difference of the sequence of Zeno leave to paradox, and harmonic, in just 4 terms pass the 1, like no paradox. But both, and all, have the paradox of the process of the infinite points (infinitesimals without measure) between 0 and 1. What you well said is the tool of integration give the answer. As dx=v.dt, integrating space-time points with dx between 0 to 1 and dt from 0 to finite T (with v out of integral, by our convenience constant) gives 1=v.T all finites really. Is a working but paradoxical tool, if I think about it. But the sequences of Zeno and harmonic are Series which, worst than integration, only sum values of distances (infinite points each one), without saying how.
The idea is to show that you can construct a sum of infinitely many halves, and then claim that the harmonic series is larger than that sum. If the first sum diverges, then so does the whole sum. So, provided that x lies between 2^n and 2^n+1, we claim 1/x > 1/2^n+1, and this lets us group n+1 terms to get 1/2.
I just played it out in my head and I think it’s kinda impossible. It’s just gonna look like this regardless of the order of magnitude on the axis: | 🙃
H(15092688622113788323693563264538101449859497) = 100.00000000000000000000000000000000000000000000900432.... It is the first harmonic number that is larger than 100.
@@nsq2487 convergence and divergence is a relative property,1/x^2 for larger values of x is way smaller than 1/x and hence the former converges while 1/x diverges.
you get infinitely many partial sum terms that are at least as big as 1/2 by following the pattern shown in the video, ie doubling the number of terms, which are all at least as big as the last one
One of my favorite youtube videos... really highlights the insanity of math and the universe, really.
I've had this question for a while. Thank you for this simple but thorough proof. Now I can apply the same methodology to understand why alternating harmonic series converge.
I am not a mathematician but this is great for me to know as a musician as its pre "tempered" - Thank you so much !
Very intuitive explanation. Thank you!
Great explanation professor!
Marvellous lesson.! Very interesting.
Very clear! Thank you
*Important question…*
In this video, all the denominators in the series are put to the power of 1, and the series diverges. HOWEVER, in the Basel equation, all the denominators are put to the power of 2, and the series converges. *question:* As the denominator powers transition from 2 down to 1, at what threshold power value will the series likewise transition from converging to diverging?
Great video!!!
Why the divergent harmonic series cannot proof Zeno Paradox of movement? Why is proved with the series of 1/2^n and not with that of 1/n if both sequences tends to 0 ? Ok series 1/2^n is convergent and our harmonic divergent, but if Zeno asked for 1/2, 1/3 ,1/4, 1/5 ,...,1/n to the destination, this divergent is one case Against the other convergent used to proof of movement. Is not a "math cheat" choose one sequence convenient to the proof without explain why others are not valid AGAINST that choosen? If the answer is only :"divergent is infinite sum" this gives reason to Zeno, as he have one valid infinite sum against that convergent proof. Who wins? I think have math answer to this, but post my initial fair though. PS: i see in the convergent as in each step summing 1/2^n to the acquired also remains 1/2^n to destination, as in the harmonic summing 1/n to the acquired remains (1-1/n) to destination.
I'll assume you're talking about Zeno's most popular paradox, _Achilles and the tortoise_
You can cover a distance of 1 by using _finitely many_ terms of the harmonic series (even excluding 1), in this case 1/2 + 1/3 + 1/4 > 1. Zeno considered _infinitely many_ steps of 1/2 + 1/4 + 1/8 + ... length to cover the distance of 1. He argued it'd take an infinite amount of time to complete this process. The paradox is resolved when we observe the time it takes to complete a step also decreases geometrically: 1/2 + 1/4 + 1/8 + ... units of time (assuming constant velocity, of course); therefore, a distance of 1 will be covered in 1 unit of time.
@@urthoperator3126 Thanks, you are right. What I said is about the real line being the same, the difference of the sequence of Zeno leave to paradox, and harmonic, in just 4 terms pass the 1, like no paradox. But both, and all, have the paradox of the process of the infinite points (infinitesimals without measure) between 0 and 1. What you well said is the tool of integration give the answer. As dx=v.dt, integrating space-time points with dx between 0 to 1 and dt from 0 to finite T (with v out of integral, by our convenience constant) gives 1=v.T all finites really. Is a working but paradoxical tool, if I think about it.
But the sequences of Zeno and harmonic are Series which, worst than integration, only sum values of distances (infinite points each one), without saying how.
In the line where you take lim on S_2^n , sign should be >=
lim (S_2^n) >= lim (1+n/2)
Thank you for this titural
Is harmonic series related to harmonic analysis?
Probably
no, harmonic analysis has to do with fourier transformatiojs
Needing 2^n terms to increase the sum by n/2 looks to me like the series diverges logarithmically.
1:48 why do we replace the 3 and 4 with 4s and 5 6 7 with all 8s
The idea is to show that you can construct a sum of infinitely many halves, and then claim that the harmonic series is larger than that sum. If the first sum diverges, then so does the whole sum. So, provided that x lies between 2^n and 2^n+1, we claim 1/x > 1/2^n+1, and this lets us group n+1 terms to get 1/2.
We can reach infinite number, then why we take 1/infenity =0. Rather we should say approaches zero. There are some maths errors.
Good one
You should show a graph of its divergance.
I just played it out in my head and I think it’s kinda impossible. It’s just gonna look like this regardless of the order of magnitude on the axis: |
🙃
it looks like the natural logarithm
H(15092688622113788323693563264538101449859497) =
100.00000000000000000000000000000000000000000000900432....
It is the first harmonic number that is larger than 100.
It's maths inefficiency to prove it convergent. Intuitively it's converges
The Harmonic series never gets to specific number. You can always add a very tiny amount and change its value an infinite number of times.
But then why doesn't that argument hold for 1/x^2? I can also add an infinitesimal value to whatever sum I have, yet 1/x^2 converges
@@nsq2487 convergence and divergence is a relative property,1/x^2 for larger values of x is way smaller than 1/x and hence the former converges while 1/x diverges.
@@aravindhsm1287 what does that even mean? Not really explaining anything
👏👏👏
Great video, but is he trying to talk as fast as humanly possibly? Had to play it in 0.9x speed
0 dislikes🔥🔥
U ruined it 😒
@@headyshotta5777 it wasn`t me! it was someone who didn`t like 0 dislikes or didn`t like the video.
Really not explaining anything here. Trying using a whiteboard or explaining what you are doing. I'm not following.
you get infinitely many partial sum terms that are at least as big as 1/2 by following the pattern shown in the video, ie doubling the number of terms, which are all at least as big as the last one
He’s explaining just fine.
It would unfortunately require some previous context
if you'd like
Hn is approx ln(n) + 0.577 for large n
Where Hn is the nth partial sum
So, if you wanna find Hn = 100
n is approx e^99.423
which is big
He's done a brilliant job