The passion you have for your subject is what makes you one of the best teachers. I wish one day I could teach with the same passion and energy. Stay blessed and thank you!
You are absolutely inspiring and so bubbly. I am a maths tutor and am studying applied maths and you're means energy of sharing your maths abiliting is so amazing!!!! Thank you!!!
I love Eddie's accent, that coupled with his energy makes want to listen to him. I get why a harmonic series is divergent from this explanation, but if it's going to infinity and infinity gives you forever to get anywhere, why then isn't a geometric series with |r| < 1 convergent? By this logic wouldn't it also be divergent?
I've got a question. The harmonic series is the sum of 1/k, from k=1 to infinity. You have just proven that it is divergent. But how about the sum 1/(a+b*k), from k=1 to infinity? At first, I thought it was the definition of an harmonic series, but I'm not so sure. Therefore, I've had trouble to find if every sum 1/(a+k), from k=1 to infinity can be either convergent or divergent. Can you help me there, please?
There's something odd on his proof.. he said that the second equation "MUST BE" smaller than the first as it approaches infinity... how does he know? it there a proof that it's smaller as the denominator approaches infinity?
So if you add an infinite number of positive numbers, the sum is always infinity regardless of how small they are. Basically, infinity = infinity, but with some steps.
I used a for loop to calculate the value of this series and turns out it approaches .693. For a while, I looked at it, thinking I have seen this number many times before only to realize its ln2. It was fun to run a for loop for 10 million times though.
I understand that harmonic series diverges because as Mr. Woo said if we let it go till infinity it goes forever, but why doesn't that happen when p>1. Thinking logically if we let it go on forever the sum should also go forever right i.e. diverge; why is that not true?
You may have already gotten your answer for this but an easy way to prove this is by taking an improper integral from 1 to infinity of 1/x which would give you ln(infinity)-ln(1) which is infinity when you try this with an improper integral with the same bounds of 1 to infinity of 1/(x^1.0001) you get {[1/(infinity^0.0001)]-[1/1]} which is equal to -1 , huh I got a negative number of well I must have done something wrong but the point is still there if you go any miniscule amount over 1 of p you will get a finite answer and for any p under 1 we can use a direct comparison test to show that since 1/x is smaller than let's say 1/(x^0.1) and 1/x diverges the former series has to diverge, hope this helped. Ps. I started the bound at 1 because at 0 you divide by zero.
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.
For people that were thrown by this as I was: I don't think you were correct to say at the end there that "the harmonic series, as it turns out, is exactly the same as the log curve and is only off by a constant". This is clearly not true since 1-log(1)=1, (1+1/2)-log(2)=0.8.., (1+1/2+1/3)-log(3)=0.7.. etc, the difference is getting smaller therefore not a constant. I think what you meant was that this difference approaches a constant as n tends to infinity, namely the Euler-Mascheroni Constant of about 0.577. Secondly you say "in fact that's one of the definitions of the logarithmic curve, it comes from the harmonic series. This isn't just like it, it IS a logarithmic function, which is crazy" which again I'm not sure is strictly true. The nearest thing I could find to a definition of the natural log that uses anything close to the harmonic series is ln((1+x)/(1-x)) = 2(x+x^3/3+x^5/5+x^7/7...) for |x|
taylor8294 he has made the harmonic curve go through the x-axis,do you thing it's correct to do so??I think the harmonic curve begins from (1,1)on the plane. Thanks!
2 (1/4) = 1/2, but the 1/4 1/4 is still there. 4 ((1/8) = 1/2, there are 4 1/8s there. This new sequence converges to zero and gives no information about the harmonic series diverging.
The harmonic series diverges because it does not converge to a specific number, not because it is incorrectly summed to infinity. Nicole Oresme was incorrect in his summation, do not continue spreading his error.
"If you add smaller and smaller values, for sure it's not going to diverge"
Harmonic Series: *am I a joke to you?*
The passion you have for your subject is what makes you one of the best teachers. I wish one day I could teach with the same passion and energy. Stay blessed and thank you!
You are absolutely inspiring and so bubbly. I am a maths tutor and am studying applied maths and you're means energy of sharing your maths abiliting is so amazing!!!! Thank you!!!
Thank you and Good Luck!
I love Eddie's accent, that coupled with his energy makes want to listen to him.
I get why a harmonic series is divergent from this explanation, but if it's going to infinity and infinity gives you forever to get anywhere, why then isn't a geometric series with |r| < 1 convergent? By this logic wouldn't it also be divergent?
Super explanation, thank you
Brillantly explained!
This video was very helpful!
Beautiful
How many ever proofs i see for harmonic is divergent doesn't convince me
For real though, I am binge watching/reading proofs right now, still don't want to believe it
@@ninakuup21 watch in numberphile channel im sure you'll belive
@@user-jx3ji9vl3z watch the proof in numberphile channel it is pretty clear.
what is the sum of 1/infinity+1/infinity+...... = ? is it convergent or divergent?
I've got a question. The harmonic series is the sum of 1/k, from k=1 to infinity. You have just proven that it is divergent.
But how about the sum 1/(a+b*k), from k=1 to infinity? At first, I thought it was the definition of an harmonic series, but I'm not so sure. Therefore, I've had trouble to find if every sum 1/(a+k), from k=1 to infinity can be either convergent or divergent. Can you help me there, please?
You should learn about the comparison test for convergence and divergence
There's something odd on his proof.. he said that the second equation "MUST BE" smaller than the first as it approaches infinity... how does he know? it there a proof that it's smaller as the denominator approaches infinity?
Integral Test for Convergence/Divergence?
I want such a teacher in Mah college!! Hey Ginnie r u listening??
So if you add an infinite number of positive numbers, the sum is always infinity regardless of how small they are. Basically, infinity = infinity, but with some steps.
I used a for loop to calculate the value of this series and turns out it approaches .693. For a while, I looked at it, thinking I have seen this number many times before only to realize its ln2. It was fun to run a for loop for 10 million times though.
The harmonic series? The first term is 1, so the sum cannot be less than 1. Not sure what your talking about.
@@FrigidOven he is talking about 1/(n+1)+1/(n+2)+.......1/(2n)
That's the alternating harmonic series- because that is the expansion of ln(1+x) except you put x = 1, which makes it ln2
I understand that harmonic series diverges because as Mr. Woo said if we let it go till infinity it goes forever, but why doesn't that happen when p>1. Thinking logically if we let it go on forever the sum should also go forever right i.e. diverge; why is that not true?
You may have already gotten your answer for this but an easy way to prove this is by taking an improper integral from 1 to infinity of 1/x which would give you ln(infinity)-ln(1) which is infinity when you try this with an improper integral with the same bounds of 1 to infinity of 1/(x^1.0001) you get {[1/(infinity^0.0001)]-[1/1]} which is equal to -1 , huh I got a negative number of well I must have done something wrong but the point is still there if you go any miniscule amount over 1 of p you will get a finite answer and for any p under 1 we can use a direct comparison test to show that since 1/x is smaller than let's say 1/(x^0.1) and 1/x diverges the former series has to diverge, hope this helped.
Ps. I started the bound at 1 because at 0 you divide by zero.
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.
For people that were thrown by this as I was: I don't think you were correct to say at the end there that "the harmonic series, as it turns out, is exactly the same as the log curve and is only off by a constant". This is clearly not true since 1-log(1)=1, (1+1/2)-log(2)=0.8.., (1+1/2+1/3)-log(3)=0.7.. etc, the difference is getting smaller therefore not a constant. I think what you meant was that this difference approaches a constant as n tends to infinity, namely the Euler-Mascheroni Constant of about 0.577.
Secondly you say "in fact that's one of the definitions of the logarithmic curve, it comes from the harmonic series. This isn't just like it, it IS a logarithmic function, which is crazy" which again I'm not sure is strictly true. The nearest thing I could find to a definition of the natural log that uses anything close to the harmonic series is ln((1+x)/(1-x)) = 2(x+x^3/3+x^5/5+x^7/7...) for |x|
...there is a book called, "Gamma."
Hi, do go on?
taylor8294 he has made the harmonic curve go through the x-axis,do you thing it's correct to do so??I think the harmonic curve begins from (1,1)on the plane. Thanks!
Teachers: Wikipedia is fake news
Also teachers: *writes the exact same thing from wikipedia*
0:44 THAT IS NOT THE CASE WITH ALL SERIES FOR EXAMPLE THE HARMONIC SERIES DIVERGES BUT THE TERMS ARE STILL GOING TO 0.
please watch the video again.
Brilliant
This 'proof' is total bullshit.
Some of his words sound British, some Aussie, some, Hong Kong accent, some American. What's going on here?
He's Australian.
american? theres none.
2 (1/4) = 1/2, but the 1/4 1/4 is still there. 4 ((1/8) = 1/2, there are 4 1/8s there. This new sequence converges to zero and gives no information about the harmonic series diverging.
What do you think is the problem in the proof provided in the video? It looks sound to me.
The harmonic series diverges because it does not converge to a specific number, not because it is incorrectly summed to infinity. Nicole Oresme was incorrect in his summation, do not continue spreading his error.