This is never an integer!

[@prunodagen]

Another way, more natural, is to show there exists a sequence of odd intengers (o(n)) and even intengers (e(n)) such that H(n) = o(n)/e(n) for all n > 1. We use strong recursion : H(2) = 3/2 = odd/even H(k) = o(k)/e(k) for all 2

[@hjdbr1094]

nice!

[@pikupal8996]

I think you should also add the proof of 1+1/3+1/5+.....+1/n=A/odd .

[@davidepierrat9072]

@@pikupal8996 the denominator would be lcm(1,3,5,7,....,n) which is of course odd

[@hjdbr1094]

@@pikupal8996 well, the denominator is a divisor of the lcm of the denominators, which in turn is a divisor of their product. All of the denominators are odd, so their product is odd and so their lcm is odd.

[@ZanderzMcCluer]

Amazing!!!

[@JalebJay]

2:24 the magic sweatshirt removal there.

[@JalebJay]

@@josephvanname3377 Interesting question

[@neur303]

@@josephvanname3377 Why do you ask?

[@jivansh]

I think m_x is not 1 but "odd". Argument is the same though.

[@mohammedsalouani7672]

Yes u r right 🤔

[@keinKlarname]

I thought the same.

[@joshhickman77]

Nah, his argument for why e.g. 3 isn't in there is good, bigger numbers end up in previous m_l entries.

[@leickrobinson5186]

@@joshhickman77 Except that isn’t m[x]. m[x] is 1 times the odd part of n!.

[@mohammedsalouani7672]

@@joshhickman77 But he multiplie by n! after.

[@CM63_France]

Hi, For fun: 4 "all the way up to", 1 "all the way up until", 1 "the first thing that I want to do", 1 "what I want to do", 1 "now what we want to do", 1 "the next thing that I want to do is go ahead and", 2 "I'm going to go ahead and" 1 "we can go ahead and" 1 "let's go ahead and" 2 "so on and so forth", including 1 "and so on and so forth", Beautifull formula : "to bring an integer to existence" 1:54 : "that a good place to ???" I can understand .

[@goodplacetostop2973]

1:51 Good place to call it ?

[@pbj4184]

You are a creative person dude. See, you can always do new stuff even when there are giants in the market. Others should learn from you instead of stealing ideas

[@goodplacetostop2973]

2:24 Shirt change 12:18 Daily homework... A tennis tournament is arranged for 2^n players. It is organised as a knockout tournament, so that only the winners in any given round proceed to the next round. Opponents in each round except the final are drawn at random, and in any match either player has a probability 1/2 of winning. Two players are chosen at random before the start of the first round. Find the probabilities that they play each other: (i) in the first round; (ii) in the final round; (iii) in the tournament.

[@pbj4184]

@Bob Trenwith I got the first one. Could you explain the reasoning behind the second and third?

[@pbj4184]

Good Place To Stop, someone overtook you today. I suggest you add the timestamp first and post the comment and then edit it afterwards to add the question. The guy who overtook you today is an Indian and KZbin takes atleast 7-10 min to get the video to us folks in India. Which means you must have typed your whole comment out before posting it. You can be first *AND* add the question if you post just the timestamp first and add and tweak the question after consolidating the 'first comment' position 😃

[@CM63_France]

2:24 : just pull out the sweet .

[@goodplacetostop2973]

@@pbj4184 Yeah I got beaten by few seconds today. It happens from time to time. It’s not a big deal because now people know me, I would be more worried about accounts “stealing” my name, my profile picture or my idea.

[@pbj4184]

@Bob Trenwith Your solution to the third one is pretty clever. Yes I too think there is a simpler solution to it

[@leickrobinson5186]

*Error* at 9:00! m[x] is *not* 1. It is 1 times the odd part of n!. [Edit: sorry, I missed jivansh’s answer 5 hours ago where they already pointed this out. :-D]

[@quantumcaffeine]

It feels like we can make this same argument more simply, using induction. Assume that, after n terms, the sum is num(n)/(2^x*odd_denom(n)), where 2^x is the highest power of 2 less than or equal to n and num(n) and odd_denom(n) are shorthand for the numerator and odd part of the denominator after n terms. Then, we add the n+1 term. The new denominator is the lcm of the old denominator and n+1, and so must be even. For the numerator, we have two cases. If n+1 is not a power of two, num(n+1) = num(n)*odd + 2^y*odd, where y >= 1. Otherwise, num(n+1) = 2*num(n)*odd + odd. Either way, we still have odd/even (with the right power of 2 in the denominator to meet the condition above). That's just a sketch, so you'll need to fill in a few details, but it seems simpler to me.

[@LucruxDCLXVI]

Mathematicians often have an aversion to doing proofs by induction

[@Vordikk]

Yes. Very simple. Came in my mind too very soon. Dislike to video, unfortunately.

[@Vordikk]

@@LucruxDCLXVI That is personal problems :D

[@andreamarino95]

There is another way. Take a prime p between n/2 and n. When you make the common denominator of the harmonic numbers, you get [ sum_{k=1}^n (n! /k) ]/n!. The denominator is divisible by p. In the numerator, all terms are divisible by p except for (n! /p) , so we get a contradiction.

[@pikupal8996]

Application of bertrand's postulate.

[@MichaelPennMath]

This was the "overpowered" method I was alluding to.

[@andreamarino95]

Ah ok sorry! I just watched the "elementary" part :)

[@ahoj7720]

There is a wonderful and elementary (though intricate) proof of Bertrand’s postulate due to Erdös using the fact that binomial (2*n, n) is an integer.

[@timurpryadilin8830]

There is another interesting result from the theory of the Egyptian fraction (fraction with unit numerator). It states that every rational number can be expressed as a finite sum of distinct Egyptian fractions. For example, 5/6 = 1/2+1/3; 17/12=1+1/3+1/12, etc. If you make a video about that, it would be great.

[@hoangkhongngo]

There is also an assertion that H(m+n)-H(m) is not an integer. Can you please show me how to prove it or introduce me a link to the proof? Thanks so much.

[@thephilosophyofhorror]

Your videos are great - keep up the good work!!!

[@suramanujan]

To be more precise, we know there exists l such 2^l

[@pratikmaity4315]

Hi Michael!! My name is Pratik and I am from India. Here is a nice Diophantine equation problem you can make a video on. It says find all a,b,c belonging to natural numbers such that a^3+b^3+c^3=a^2b^2c^2. Love from India.

[@GreenMeansGOF]

That’s creepy. I was literally thinking about this problem last night.

[@lyrimetacurl0]

Great minds think alike!

[@Oscar1618033]

Let p be the bigger prime smaller than (or equal to) n, then thanks to Bertrand's postulate we can say that n/2 < p 1, therefore p must appear only one time as a factor of all the denominators. Suppose that Hn is integer, than Hn-1/p should have p in the denominator (after reducing) implying that p appears more than one time in the sum. Absurd, therefore Hn is not an integer.

[@amaraparajeet6339]

Oh man This was a question in a college entrance exam I was preparing for

[@МаксимАндреев-щ7б]

Exist k: 2^k k>=1 -> A is even. A*1/m is even for any m != 2^k, A/(2^k) is odd. Then A*H_n is odd, but A is even -> H_n is not natural.

[@mihaigabrielbabutia4595]

Neat and very clear! Keep up the good work!

[@pierineri]

To make it simple: let 2ˣ = the maximum power of 2 less than or equal to n. All terms 1/k for 1 ≤ k ≤ n, k ≠ 2ˣ have a common denominator of the form : a power of 2 strictly less than 2ˣ times an odd number, giving a sum of the form EVEN/( 2ˣ.ODD). Summing the remaining 1/2ˣ, we get H_n = ODD/( 2ˣ.ODD). To make it even simpler: sums of fractions with denominator any odd multiple of a fixed power of two 2ᵃ, (and numerator any integer) are still of the same form. For a:=x-1,all terms 1/k are of this form but one. (here "ODD" stays from some odd number, not necessarily the same)

[@medexamtoolscom]

If I were to try to do this on my own, I would try to make an induction argument, like take the partial sum up to 1/(n-1), and separate it into 2 cases. One if the denominator of the reduced fraction version of that, i.e. the lcm of 1 through n-1, divides n, and one if the lcm of 1 through n-1, does NOT divide n. Let's see if that's what you do.....

[@cameronspalding9792]

One can find an upper and lower bound for the sum of the first n terms using integration

[@pbj4184]

Could you explain your solution?

[@onealjerry]

True, but that only works if we can show that the upper and lower bounds don't have an integer between them. I doubt this would work in all cases.

[@pbj4184]

@@onealjerry Yeah I think the upper and lower bounds are ln(n+1) and ln(n) because Hn is between them. Their difference is indeed less than 1 but that is no guarantee that they don't have a natural number between them.

[@pbj4184]

I think what the OP means to say is that we integrate 1/x from 1 to n. That will always be less than Hn. And similarly the integral of 1/x from 1 to n+1 will be less than H(n+1). The integrals evaluate to ln(n) and ln(n+1) respectively. And ln(n+1) > ln(n) clearly because ln(x) increases with x when x is positive. => ln(n+1) > Hn > ln(n) > H(n-1) We get bounds for Hn and their difference is less than 1 too* so we can't say surely that they have an natural number between them. But we can't decisively say that they don't either (EDIT: Yes there is a way to generate counterexamples. So there is nothing left to chance here. Read the addendum below) If they do, then we can't use this method to prove Hn is not natural. The OP explaining their method would clear this confusion up. * ln(n+1) - ln(n) = ln( (n+1)/n ) = ln( 1+ 1/n) The maximum value of 1/n is 1 when n is natural (And 1/2 in fact, since we consider n≥2). So ln( 1 + 1/n ) can be ln( 1 + 1/2 ) at most which is ln(1.5) which is clearly less than 1=ln(e). So the difference of ln(n+1) and ln(n) is always less than 1 for natural n EDIT: Read Jeff Zhang's comment to see how to generate counterexamples. Thanks a lot Jeff! P.S- Like Jeff's comment too, please :)

[@boredomisbliss6271]

@@pbj4184 there are absolutely n where there's a naturals number sandwiched between ln(n) and ln(n+1). Floor(e), floor(e^2), floor(e^3), floor(e^4), etc

[@kostaspapadopoulos1480]

It can be generalized that after the 2^(k+1) th term there is no partial sum with k or less decimal digits, because if there is, we first multiply both sides with 10^(k+1)=2^(k+1)x5^(k+1) and the right side of the equality becomes an even number. Then we multiply both sides with 2^(λ-κ-1) times the product of all the odd denominators, where λ>=k+1 is the exponent of the largest power of 2 which is

[@themetricsystem2899]

such a nice proof

[@wavyblade6810]

9:42 shouldn't it be the white odd number instead of 1? Anyway, the bracket will still be odd.

[@willmoller3646]

Yeah…

[@spicymickfool]

Does this work? H_n is the nth harmonic number. H_n=1/n+H_(n-1) -> nH_n=1+nH_(n-1). So if H_n is an integer, then so is nH_n and nH_(n-1). n is a common factor of both , but has to divide their difference, 1. This is a contradiction, so H_n is not an integer for n>1.

[@javieramado6839]

I love the proof by contradiction, assuming the sum is an integer. Then multiplying by the product of the odd numbers and 2^k, and showing that odd=|= even

[@boukaddid]

شكرا جزيلا أستاذ

[@baruchben-david4196]

It is if you stop after the first term...

[@jacoboribilik3253]

He said that, know-it-all.

[@ffggddss]

Technically (this IS mathematics, after all!), there are *two* values of n for which H_n ∈ ℤ. Namely, n = 0 and n = 1: H₀ = 0; H₁ = 1. Fred

[@TikeMyson69]

Indexing starts at 1.

[@ffggddss]

@@TikeMyson69 Yes. But summation from m=1 to 0 is a valid mathematical operation. It's called a "vacuous sum," and, being a sum of 0 terms, it automatically evaluates to 0. Similarly, a vacuous product automatically evaluates to 1. Fred

[@TikeMyson69]

@@ffggddssI'm trying to get my head around this definition. I'm in trouble with it because in this particular sum n=0 would make the 0th term to be 1/0. For this reason allowing vacuous terms should be part of the stipulation of the problem. Otherwise it may be considered as an abuse of notation.

[@ffggddss]

@@TikeMyson69 1/0 never enters into this. Compare this idea with the argument that 0! = 1, because, by the recursion, (n-1)! = n!/n and setting n = 1, gives 0! = 1/1 = 1 In the summation case, ∑[m=1 to n-1] a(m) = {∑[m=1 to n] a(m)} - a(n) Then, setting n = 1 gives ∑[m=1 to 0] a(m) = ∑[m=1 to 1] a(m) - a(1) = a(1) - a(1) = 0 And that's how it follows. It is no abuse of notation; it's widely accepted in mathematics. Fred

[@jarisipilainen3875]

until you reach 1/314159 its closest

[@soutriksarangi5580]

A qn based on this property came in India IMO Training Camp 2019 Practice Test-1. Nice problem.

[@felipelopes3171]

A very simple result if you understand p-adic valuations

[@kikiak4971]

i proved it in by induction as following , but i don't know whether it's correct let A=1+1/2 +1/3...+1/n if A is an integer，then A=[A] if A is not an integer，then A=[A]+{A}, {A} 0 (1) when n=2，A1=1+(1/2)=3/2,[A1]=1,{A1}=0.5 0 (2) let n=k，let [A(k)]=N,{A(k)}=A(k)-[A(k)]=A(k)-N0 (3) then n=k+1， A(k+1)=A(k)+1/(k+1) [A(k+1)]=[A(k)+1/(k+1)]=N+[1/(k+1)] {A(k+1)}=A(k)+ 1/(k+1)-N-[{A(k)}+1/(k+1)] {A(k+1)}={A(k)}+1/(k+1)-[{A(k)}+1/(k+1)] so {A(k+1)}0, actually 0

[@theo_pap_]

Sadly it is not correct. In particular, you use that [p+q]=[p]+[q], which is not true for all p,q. E.g. 11/6+1/4=25/12, [11/6]+[1/4]=1+0=1, but [25/12]=2. You cannot add the integer parts like that, because eventually adding 1/n to the previous sum will be enough to make the new sum greater than the next integer. As the harmonic series diverges, this actually happens infinitely many times.

[@kikiak4971]

thank you for pointing out that

[@senhueichen3062]

NICE finding. It diverges to infinity, so it actually passes EVERY integer.

[@joshhickman77]

*positive

[@martinepstein9826]

He wasn't taking the limit at n -> 0

[@longlostwraith5106]

No, that's irrelevant. There are infinite odd numbers but none of them is even... Just because it grows infinite, doesn't mean it can be an integer.

[@TheoH54]

And this one is, quite surprisingly, ALWAYS an integer: SUM(k=1,n) k.(1+1/n)^k (Writing it out for n=10 for example makes it clear why it's surprising - it's not difficult to proof this property but nonetheless it's 'weird')

[@Czeckie]

isn't that always n^2?

[@TheoH54]

@@Czeckie for n=1 sum=2, n=2 sum=6, n=3 sum=12 . So yes, sum = n(n+1)

[@GhostyOcean]

Induction ALMOST works, but I think it would require some extra work that I'm not gonna do in a KZbin comment. n=2 SUM(1/i) from 1 to 2 =1+1/2=3/2 which is odd over even n=k assume that SUM(1/i) from 1 to k is of the form (2a+1)/2b for integers a,b with b≠0 n=k+1 SUM(1/i) from 1 to k+1 = SUM(1/i) from 1 to k +1/(k+1) =(2a+1)/2b +1/(k+1) by induction hypothesis =[(2a+1)(k+1)+2b)]/[2b(k+1)] =[2ak +2a +k +1 +2b]/[2b(k+1)] =[2(ak+a+b) +k+1]/[2b(k+1)] here is where it fails. It works for even k, but odd k requires more work that I don't want to type.

[@weckar]

This is just a specific case of the fact that "For no set of which the unique members are in the form 1/n, does any combination of elements make an integer." right?

[@klementhajrullaj1222]

But how is nearly that sum?! ...

[@brucea9871]

I heard this statement before but never saw a proof of it. (I'm also aware that the harmonic series diverges and there is a simple proof of that if you think of the right trick; group the terms in groups of 1, 2, 4, 8, 16 (the powers of 2) and show that the sum of each group exceeds 0.5 and since the sum of an infinite number of numbers each greater than 0.5 must be infinite this shows the harmonic series diverges to infinity.) Of course a more sophisticated proof is the integral test using the function n to the power p which proves convergence if p is less than -1 but divergence if p is greater than or equal to -1. But that was a quick shirt change at 2:24. Who is this guy anyway; Quicksilver of the X-Men?

[@sidimohamedbenelmalih7133]

Pleas penn: what is all naturals numbers such that the sum of the invers of his dividers is also a natural number

[@ImaginaryMdA]

Ooooh, I took a prime p large enough so that only one term had a denominator divisible by p. This quickly leads to a contradiction.

[@Czeckie]

hmm, I have no idea what could be the "high power theorem" you are alluding to at the end

[@pbj4184]

Are you being sarcastic or do you really don't know? Please do tell us if you know!

[@saraqael.]

Prabhanjan Sahoo It’s “Bertrand‘s Postulate“. Here‘s a proof to the problem in the video using this postulate: proofwiki.org/wiki/Harmonic_Number_is_not_Integer/Proof_2

[@pbj4184]

@@saraqael. Oh I see. Thanks a lot for making the effort of searching for and linking the proof! 😀

[@medexamtoolscom]

"unless n is 1" * unless 1 is 0 or 1.

[@dennisalessandroamoruso2127]

Hello! Can I try to explain a different idea to proof this question? My idea is a proof with an absurd. OK we check that there is one integer that our argument is true. For easy example n=2. Now we accept the idea that the subset of integers with Hn is an integer is not empty. So we can take the littlest integer m that Hm is an integer. This is possible for the order of Z. So for m >2 Hm=k belongs N( Z but obviously Hm>0). Hm= Hm-1 + 1/m so Hm-1= m*Hm - 1 but it is simple to proof that Hm-1 is an integer but for our hypothesis m is the minimum integer with the property of Hn is an integer. So we arrive at one contradiction and we win! 😀 Is it a good idea? Sorry for my English but I have some problems with some words. Thanks for video and I love the idea for a discussion of my idea of proof. Thanks all!

[@ericslavich4297]

It's a good try, but... You say "Hm= Hm-1 + 1/m so Hm-1= m*Hm - 1" but I think it's H(m) = H(m-1) + 1/m implies m*H(m-1) = m*H(m) - 1. Note the factor of m on the left-hand side. And showing m*H(m-1) is an integer does not show that H(m-1) is an integer.

[@dennisalessandroamoruso2127]

@@ericslavich4297 Thank you for your Comment and I am very sorry about this silly mistake. I will try another road to proof this. Thanks.

[@dennisalessandroamoruso2127]

@@ericslavich4297 sorry but maybe I have another idea to continue my proof. We proof that m*Hm-1 is a positive integer but Hm-1 is not an integer. Ah another important relation is m*Hm-1=m*Hm - 1 and the fact that Hm is a positive integer. OK. Hm-1=a/b with a, b positive integers, GCD(a, b) =1 and b>1 because Hm-1 doesn't belong to Z. So m*(a/b) =m*k - 1 with k=Hm. So we arrive at b=m*a/(m*k-1). Now GCD(m, m*k-1)=1 and m*k-1>m because for hypothesis m>1 and k>1 because for every n Hn =1 + z with z positive number because it is the sum by 1/2 to 1/n. So b is an integer if and only if m*k-1| a. So a= s*( m*k-1) and b=s*m but GCD (a, b) =1 and so we have an absurd! Whato do you think about this idea?

[@braveCod3r]

that is a good way to stop

[@sentipy8990]

Where did jacket go? xD

[@RF-fi2pt]

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.

[@pierrot31511]

Nice video, Michael. Could you talk about the serie of the inverse of prime numbers and Brun theorem ! It would be great !

[@MichaelPennMath]

I have a video on the sum of reciprocals of primes already!! kzbin.info/www/bejne/q2TToGl7rrZ-pJY

[@pierrot31511]

@@MichaelPennMath What about Brun theorem/constant ? Is that too difficult ? Thanks for your ansmwer. Just a remark : why don't you show results about matrix and linear algebra. In my opinion, it is the only lack in your great channel !

[@roberttelarket4934]

Make sure to vote 11/3 for Michael Penn for President-the only hope for the U.S.!

[@pbj4184]

But then he won't be able to make good math videos :(

[@roberttelarket4934]

@@pbj4184: Nonsense! As a top notch mathematician Penn will make outstanding probabilistic decisions with his computer/brain. After all he solves math problems wonderfully in his videos so again equally as well he would solve other non-math problems mathematically/symbolically. He'll have plenty of time to make videos and of course leave the little details to his staff. Are you aware that Ataturk the founder of modern Turkey about 100 years ago was a mathematician?

[@pbj4184]

@@roberttelarket4934 Wow I didn't know Atatürk did math. Thanks for teaching me an interesting fact!

[@roberttelarket4934]

@@pbj4184: No problem.

[@simaomoreira20]

Road to 100k

[@factorial1059]

What's the name of theory

[@saraqael.]

It’s “Bertrand‘s Postulate“. Here‘s a proof to the problem in the video, using this postulate: proofwiki.org/wiki/Harmonic_Number_is_not_Integer/Proof_2

[@aakksshhaayy]

So obvious a postulate....

[@humbledb4jesus]

i guess 'i can see that' isn't an answer, right professor?

[@schweinmachtbree1013]

the zeroth harmonic number is an integer too ;)

[@pbj4184]

But n is a natural number :)

[@schweinmachtbree1013]

@@pbj4184 Here we go again xD

[@pbj4184]

@@schweinmachtbree1013 😂

[@riccardocaiulo8418]

👺

[@MrConverse]

Yeah, you lost me.

[@rudychan2003]

First dislike, there is answer; Approximate logarithm ➡ H(n) = ln (n) + T* T = 0,577...~ 🙄 Collecting data: n = 100, 1000, 10.000 and so on.! 💹

[@pbj4184]

Wow this is sooooo rigorous

[@pbj4184]

And in case you didn't know, irrational + irrational can indeed be natural

[@pbj4184]

First dislike on your comment

[@mr.cheese5697]

Is this needs to be proved? It's obvious.

[@jacoboribilik3253]

Indeed. I was dragged out of the womb knowing this and a couple Tauberian theorems.

[@rateeshk8175]

Blackpenredpen already did a video Too late

[@garvittiwari11a61]

First

[@quirtt]

Do I know you?

[@garvittiwari11a61]

Yep runtime_error on tele

[@pbj4184]

Agle baar nahin ho paoge. Enjoy it while it lasts 😉 (Translation: You won't be first again. Enjoy it while it lasts)

[@pbj4184]

@@garvittiwari11a61 No offense by the way. It's just that I want Good Place To Stop to be first :)