my new favorite proof of this classic result
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- Опубліковано 1 кві 2024
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You know you're early when good place to stop is not here lol
Someone's future book: "and that's a good place to stop" a biography on the life and work of Michael Penn.
So, we know how it ends. "Michael died. And that's a good place to stop."
I think that will (in a hopefully very, very far future!) be Michael's famous last words!
@@talastra Maybe he won't ever die. You never know. But regardless, the last sentence MUST be "and that's a good place to stop."
Since The good place to stop man is late, I would like to help. Good place to stop: 11:20
good thing I checked the comment section before 11:20 and saw this. Thank you!
You can also Fibocrash the thing by pointing out that (Fn/Fn+2) = (Fn/Fn+1)(Fn+1/Fn+2) and this approaches 1/Φ^2, hence the general term is not infinitesimal.
Once you establish that the harmonic series is >= 1+Sum(Fn/Fn+2), the way I intuitively understand it is that the series terms will gradually approach 1/1.618...^2 because the fibonacci sequence approximates the golden ratio. I know that's not as mathematically rigorous, but I think it would make the whole thing easier to grasp depending on what level your listener is at math-wise.
That is definitely the right intuitive perspective, which Michael could have mentioned. But then you need to work that into a proof, which he did quite neatly.
The one after the next F-number: Easier to say in german „übernächste“ F-number. Nice proof, thank you, Michael.
It is great that you showed this nice trick. Thank you always for your amazing presentations. Thank you.
Love this. Math has a beauty that is hard to explain
Some intuition for the inductive hypothesis: notice that each subsequent term in the Fibonacci sequence is *almost* double its last term but not quite. So two terms ahead is almost four times larger than its current term, which is certainly less than five times larger than its current term. Hence we can safely say that the ratio of the current term to the term two steps ahead of it is at least one fifth. Infinite fifths is infinity and we’re done.
The five is very weird nonetheless. It is pretty easy to see that any factor works if (and only if) the base cases work. So if the statement would be done with "less or equal" instead of "less", the factor 3 would work (and even be optimal), even if you want to keep the strict less and the factor a natural number the factor 4 works. But pi works too! So why the hell 5?!? I could understand choosing 42 but not 5...
Where is good place to Stop
when the world needed him most, he vanished
@@luckycandy4823
He will return.
there was none. it's a divergent video
11:20 ???????
Yeah I’m late, I’m sorry 😢
Very nice! Thank you!🙏
I really like the video, simple and elegant.
I'm pleased to report that this lecture -- like his video on multiplying matrices -- made no sense whatsoever. So, another growth opportunity! I'll do some studying and get back to this.
Why doesn't it make sense according to you? I'm asking because I think I had the same thought as you but mine was incorrect. I'm curious what you think, maybe I've missed even more than I thought.
very smart solution!!
Actually, the Fibonacci sequence is almost geometric progression. You can prove the divergence not only like 1+1/2+1/3+...>1+1/2+2*1/4+4*1/8+...=1+1/2+1/2+1/2+... but like 1+1/2+1/3+...>1+1/n+1/n+...+1/n (n-1 times) + 1/n^2 + ... + 1/n^2 (n^2-n times) + ... = 1+(n-1)/n+(n-1)/n+... with arbitrary n>1.
Why use a factor of 5, won't a factor of 3 work just fine?
Though I suppose you have to make it ≤ in that case. 5 is kind of nice because of its relationship to the Fibonacci numbers though :)
He said the inequality is not tight
anything >or= 3 would work just as well, like pi
I have clearly found the right place on UA-cam, or been influenced watching Prof Penn’s videos the same way as others, as I also thought “why not use 3?” Then “anything over 3 works…how about pi?”
Yes, 3 would work instead of 5 as long as what we want to prove is changed from Fₙ₊₂
this is a really beautiful proof
thank you for sharing it -- it brightened my day
I feel like one method is shown way too much on UA-cam. This is nice to see
Great Prof
That is strange -
you would think that the series would converge on a final value but it doesn't.
any autoregressive sequence (such as fib) is asymptotically exponential, so this represents a very small change from the standard cauchy condensation test.
Agree. this really isn't that different from the standard way of showing the harmonic series diverges. Indeed as you sort of indicate, you can generalize this technique to cover both
Perhaps you can notice that (F(n)/F(n+2)) converge towards 1/φ^2…
I think my favorite is by Cauchy condensation test, but this is interesting. Once you've estimated the series in terms of a sum of ratios of Fibonacci numbers, you can bound this by a (divergent) geometric series, using well known formulas for Fibonacci numbers (golden ratio!). Sometimes you should do the asymptotic series (in terms of Bernoulli numbers) of the Harmonic numbers (partial sums).
The Cauchy condensation test could have a variant with the Fibonacci numbers instead of 2^n, if only it weren't so silly... ;>
amazing!
Michael was right. This was fun.
I like replacing the F_n/F_(n+2) successively with F_n/[F_n + F_(n+1)] and F_n/[2F_n + F_(n-1)] by the definition of the Fibonacci sequence itself. Then replacing the F_(n-1) term with F_n makes the denominator strictly larger for all non-trivial values of n, thereby making the fraction smaller. Now reduce to a simple 1/3. The Fibonacci ratio series is then bigger than yet another divergent series, and we're done!
Much, much simpler is the following:
Assume S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + .... is finite. Then
S > 1/2 +1/2 + 1/4 + 1/4 + 1/6 + 1/6 + .... = 1 + 1/2 + 1/3 + ... = S.
Contradiction.
This isn't about the simplest proof but a fun proof.
@@NotBroihon The proof presented is basically the same idea as the old proof grouping by powers of two.
The proof given is nice enough but I will not really categorize the proof given as particularly fun.
Suppose the harmonic series converges to L. Then
L
= 1+ 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + …
> 1/2 + 1/2 +1/4 +1/4 +1/6 +1/6 +…
= 1 +1/2 +1/3 +…
=L
Thus L>L, a contradiction. So L does not exist.
I don't see a contradiction here. On the left, the sum has more terms than the sum on the right for each step. Of course, there's an inequality >.
(I know, ofc, other proofs for the harmonic series that are valid.)
Very neat!
@@Achill101 there are the same number of terms in both series. A series must either be finite or have countably infinite terms. Neither is finite so both are countably infinite.
@GhostyOcean - the limit of the series of a sequence (a(n)) is defined as the limit of the partial sums, s(m), with s(m)=Sum(1...m, a(n)). For the partial sums you written up, the inequality holds (because the lower one has fewer terms), which doesn't automatically mean the limit wouldn't exist.
@@Achill101 once again, they have the same number of terms. I agree that the partial sums are NOT the same, but to say there are a different number of terms is demonstrably false.
Amazing proof. I really enjoyed this vid; reminds me of why I subscribed to the channel in the first place. Solid logic, interestingly creative proof of something we know to be true using other things we know to be true. Such is math 😌
I think the end of the proof could be easier. Rather than use induction, simply replace F_n with F_{n+2} - F_{n+1}, which is true by the definition of the Fibonacci sequence.
Then you have F_{n+2}/F_{n+2} (which is obviously 1) minus F_{n+1}/F_{n+2}, which never goes above one because the previous Fibonacci number is always less than the next one (and as n --> inf, the ratio approaches 0.618... by another well known property of the fibonacci sequence).
And that's it. You're adding up infinite positive terms that don't go to zero.
I think that for the sake of having a self-contained proof, Michael did not want to use the well-known limit property of the series... Otherwise you could also use F_{n}/F_{n+2} > F_{n}/F_{n+1} -> ~0.618
He mentions he is not using that limit for the purpose of the proof being more self contained
That inequality is wrong. @@antonionanni5893
That inequality is wrong, it does not work. @@antonionanni5893
@@antonionanni5893 You need 1/φ², not 1/φ.
Very interesting explanation. But what was the significance of 5? I mean, if we were to prove F_(n+2)=3 would do. This really shows how the harmonic series fails so badly to converge that many trivially divergent series could be readily found which are strictly less than this series when truncated.
He literally said the inequality isn't tight, please pay attention to the video
Could someone please tell me why you need two base cases in the induction? Is it so you can justify using two induction hypotheses?
I find it very interesting that we can say that the harmonic series is greater than 1 + the infinite series where every term is 1/5
Really nice one!
Why not instead prove that F(n+2) < 3F(n) for n > 2
For n > 2, F(n) > F(n-1), and F(n+2) = 2F(n) + F(n-1) < 3F(n)
He literally said the inequality is not tight
What is the boundary (for n) where the sum of 1/x^n converges - given 1 doesn't and 2 does? Pondering but can't quite see!
It converges when n > 1.
@@evgenymakarov8605 Thankyou! :)
Before watching this video, I thought it's about Fourier transform of the Fibonacci closed form formula into a series of sines and cosines. Can that be done?
Could we have said that Fn/Fn+2 converges into 1/phi^2, so that summation diverges.
Idk how you'd define this but it's a very intuitive leap of logic from that step to our conclusion.
That's right, by the limit test. And when a series with positive terms is divergent, it diverges to infinity.
is the Integral test good place to stop ist it
Good video.
wow Fibonacci is everywhere
I'm lost. Why did we place a Fibonacci number in the numerator?
Lets take the third group as an example (1/4+1/5). You have F3-terms that are bigger than or equal to 1/F5, so when you make the replacement you get (1/4+1/5) >= (1/5+1/5) = 2/5, but the 2 in the numerator is F3 and the 5 in the denominator is F5, so you get (1/4+1/5) >= F3/F5. So for each group you're replacing each of the Fn-terms with 1/Fn+2, which add up to Fn/Fn+2
@luisaleman9512 Oh I see, thanks!
ok, but is there a similar result for the sum of all fibonaccis like the -1/12 is for the naturals ?
Put it in the ramanujan summation formula and look what happens. Probably some ugly unsolvable integral lmao
Assuming I entered everything correctly I got approximately 0.0433546 + 0.0140207i. Not nice.
can you make a video on the famous sum of the natural numbers = -1/12 and why this may deceive many students?
It is only true if you change the definition of the infinite series. It's hard to present it as a universal truth.
Just wonderful ❤
Please Integrate (x^2+5)/(x+2) dx with Algebraic Manipulations and explain. Using Algebraic Identities would be better.
(x^2+5)/(x+2) = x-2 + 9/(x+2) and the rest should be quite simple.
confused again
Is it just me or is the volume a bit too small in this video?
What is not clear is why there are F_n fractions in a "group" of fractions?
because the F_(n+2)-th term minus F_(n+1)-th term will remain F_n number of terms, so each part groups with that many number of fractions. (I’m not quite sure what you are asking.hope i answer what you want.)
that's just the way he decides to group them. The real question is why each fraction in the group n is bigger than F_n+2, ie why is 1 + F1 + F2 + ... + Fn = F_(n+2). but this can be proved with induction :
if this equality hold for a number n, then :
F_(n+3) = F_(n+2) + F_(n+1)
and by induction hypothesis
F_(n+3) = 1 + F1 + F2 + ... + Fn + F_(n+1)
CQFD
If after a finite number of terms, all terms bear the same sign, the behavior at infinity is independent of the grouping chosen
@@Joffrerap No, your first question occurs because the given explanation in the video is incomplete. If you decide to group all terms between 1/F(n+1) (not included) and 1/F(n+2) (n>=1) it's always possible because the sequence (F(n)) is never constant from n=2. In each group the smaller is the last. The number of terms in each group is F(n+2)-F(n+1) but F(n+2)=F(n+1)+F(n) thus F(n+2)-F(n+1)=F(n). These are the missing explanations.
@@ojas3464 I know that but it's not really related to my question. My english is bad but i haven't seen written and justified why there are exactly F(n) integers in the interval ]F(n+1),F(n+2)].
Lovely proof! I like the blackboard staging... Makes me feel young again...
I dont understand? How can F3 be equal to 2 AND be equal to 1/4 + 1/5 ? It doesnt make sense.
F₃=2. It doesn't equal 1/4+1/5, rather it is the number of terms summed (2). 1/4+1/5>1/5+1/5=2/5=F₃/F₅
Who draws an 8 like that? Two circles? It’s an upright infinity symbol. No need to remove your chalk from the board.
End result of 1 + 1/2 + 1/3 + 1/4 + 1/5 + ...... > 1 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + ........ does not seem to be correct both intuitively and according to the python script I ran for n = 10000
In the inequality, there are many more terms on the left than on the right.
. . . Similarly to the standard proof, where you replace 1/n with 1/2^m where 2^m is the next number larger than n of the form 2^m. Then you get
1+1/2+1/3+1/4+1/5+1/6+1/7+1/8
> 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8
>
1+1/2+1/2+1/2
The inequality is true, because there are 8 terms left of > and only four terms to the right of it.
You probably counted the terms incorrectly. You need to compare the sum of the first Fₙ terms of the first series with the first n terms of the second series (roughly speaking).
the best proof ever is the following (don't argue with me)
we know that H_{n} is strictly increasing so its limit is either a finite reel number or infinite.
We have for all n, H_{2n}-H_{n}>=1/2. so of we suppose that the limit if finite reel number L then by taking the limit we will get :
0= L-L>=1/2 which is wrong.
Therefore the limit of H_{n} is infinite
The authors of this proof are to be congratulated on spotting the Fibonacci nos lurking in the humble harmonic series, but methinks (a cod medieval word) nothing will ever to Nicholas Oresme's classic ~1350 proof - a model Mathematical insight
I beg M. Oresme's pardon; its _Nicole_ Oresme. See here:
en.wikipedia.org/wiki/Nicole_Oresme
Ehhh the Fibonacci sequence is superfluous here. Any exponentially growing sequence works, and the necessary lemma is more immediate for 2^n.
cute
I'm sorry but I think we all agreed that this is clearly equal to -1/12?!!!!!
No, it’s the sum of all natural numbers (1+2+3+…) that is somehow related (via analytic continuation in complex analysis) to -1/12
Your plausibility argument is not a proof that the harmonic series terms will always be greater than this ratio of Fibonnacis. Since the denominators are simply counting, adding a block of Fn terms after a term with denominator Fn-1 always yields a last term with denominator F_{n}+F_{n-1}=F_{n+1}. Since the smallest term in a block with Fn terms is always the last term with denominator F_{n+2}, the sum is always greater than (or equal to) the number of terms times the smallest term...qed