This Problem will Fool You 

🎓Become a Math Master With My Intro To Proofs Course! www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C

Understood every step ✅ Makes sense to do it this way ✅ Could have dreamed up the solution ❌ Not today, maybe in my youth. A troll of a question to put on an exam.

I like how he "researched" For a while through yt polls, and now is constantly bringing these great videos

I do not follow how the 1/sqrt(k) inequality at about the 6 minute mark implies the a_k inequality. The inequality on the cubic term goes in the wrong direction due to the negative sign.

great video!! ive always struggled with induction, i like how much you hammered the steps in, i really started to get it

I wonder if there's a power we could take a_n to to make the series converge

5:56 Awesome video! Just a minor question. What justifies the last line? Since a_k >= 1/sqrt(k), we can replace a_k with 1/sqrt(k) in the first term. But we cannot just do the same replacement for the second term since we are subtracting.

I saw the sine, and it opened up my eyes, and it’s also good to have your W.T.S. about you. Thanks for another wonderful video!

I feel like home when I watch these videos

Loved this one. If you mix up the strategy a little, you can prove that the series sum for (a_n/log(n))^2 does actually converge.

Keep going mate, this channel is destined to grow.

Great video as always

Besides being good, when you squared the inequality you made both sides positive so the inequality may not work, e.g. -2^2 > -1^1 but -2 < -1. If both sides must be positive then it won't matter.

This one made my head hurt. You lost me when I realized you weren't talking about Sinex, the nasal spray. It seemed the approach of comparing to the harmonic series wasn't terribly intuitive, predicated as it was on the assumption that the series would diverge. If you first assumed otherwise, you'd spin your wheels for quite a while trying to find a way to tackle the problem. I'll definitely have to watch the video again.

The terms a(n) which are squared in the series are generated through a ricursive formula a(n+1)=F(a(n)) , a(0)=π/2, where the function F=Sin has a "contractive" behaviour, since its derivate Cos has absolute value strictly minor than 1(only for 0 argument Cos(0)=1, but 0 is the fixed point, i.e. Sin(0)=0 and this would definitively set the sequence to zero, vfr infra). As a consequence we can for instance apply the Caccioppoli-Banach Theorem and so we can rigorously prove that the terms a(n) tend to zero as n tends to infinity. In fact, whenever the CB theorem can be correctly applied, the limit of the sequence a(n), as n tends to infinity, is the so-called "fixed point" a that verifies the equation a=F(a); in our casa F=Sin, and so we obtain the equation a=Sin(a), whose only solution is a=0. The abovementioned proof is omitted in the video,(replaced by an implicit intuitive claim?) Obviously, the zero limit of the terms is only a necessary condition for the convergence of the series. In fact, the actual divergent behaviour of the series is rigorously studied in the video by means of comparison of terms with the well known harmonic series; the approach is cleverly based on induction, and this is coerent with the nature of the ricursively-defined sequence. The convergence behaviour of infinite series which may be built with higher power of a(n) can be easily studied through the generalization of the described method (a claim of comparison with a known series to be inductively proved;Taylor approximatio approximation of Sin...)

Love your videos 🤗🤗

Summations actually are deceiving. Their answers are so... so gorgeous. Wonderful that adding and subtracting a bunch of things gives you an elegant solution

Take the sum of something squared. Something squared is always positive. The sin function cycles, so nothing decreases faster than it's being added as N trends towards infinity => the series diverges.

Was able to follow, but there is no way I would be able to solve this alone.

Oh this seems fun. Since sin x is less than x for positive x we know the sequence is always positive and decreasing. Limit of ratio might get weird though since (sin x)/x approaches 1. Should be a fun video.

Thank you Professor Brian Hawthorn aka Brithemathguy!

I came up with a different proof. Since a_n goes to 0 we can make a_n small enough so that a_n > a_(n+1) = sin(a_n) >= a_n - a_n^3/6 Now think about how many terms it will take for a_n to half. For a conservative approximation suppose that a_(m+1) is always a_n^3/6 less than a_m. Then it will take at least [a_n/2]/[an^3/6] = 3/an^2 terms to get to an/2. In this time we always are adding at least (an/2)^2 to the sum. So before halving a_n we need to add at least 3/an^2 *an^2/4 = 3/4 to the sum. Since we can half infinitely many times, we add 3/4 at least infinitely many times, so the sum diverges

Wow!

I think there's a mistake with the minus sign in the inequality in 5:57.

I understood everything up until he said that because k=1/3, 1/8 that it implies that the induction hypothesis is true for all natural numbers. I don’t understand how two fractions prove that it works for all natural numbers

Well the problem did fool me; I would have thought that sin(a_n)^2 was decreasing fast enough to converge, because sigma(1/x^2) converges.

Nice video :-)

This guy casually pulls out gamma functions out of his pocket but he has to explain a simple method of proving stuff.

you kind of messed up your notation when doing the algebra part. you used implication arrows, but obviously you would want to use equivalences, since you started with the claim and ended with a true statement.

Damn I fell for it too

It sgoudent be by the limit test no conclusion

Papa Flammy is shaking.

5:28 "If we can show [a sub k minus a sub k cubed over three factorial was greater than one over square root of k plus one], then that would [imply our previous step, which is what we want to show]". Me: (almost catching up) 5:54 "Now we want to show this: [one over root k minus one over root k cubed over three factorial is greater than one over square root of k plus one]. It's a bit of a mouthful and a bit clunky, but if we prove this, that would [imply our previous step which would [imply the previous step, which is what we want to show]]." Me: uh… um… is this the domino effect of induction?

hmmm These problems are very interesting although it does seem mean to put this on a test

The lines of reasoning for these problems are always so unintuitive. You'd have to suspect that the series diverges to even consider comparing to the harmonic series. And choosing the truncation of the Taylor series seems kinda arbitrary until you see it work out later. It amazes me that people can somehow tackle these problems under time constraints, even if this is one of the hardest math exams out there.

At 5:04, what's W.T.S?

😄👍

Jokes on you mate, all problems fool me.

Let's Consider the ratio test: If the following inequality holds then the given sum is absolutely convergent limit k->inf [(sin( ...(k+1 times)...(1))/sin(..(k times)..(1))^2] < 1 Base case: (sin(sin(1))/sin(1))^2 ~ 0.785 < 1 OK. Induction step: Consider, bwoc, there exists some n such that (sin( ...(n+1 times)...(1))/sin(..(n times)..(1))^2 >= 1, then (sin( ...(n+1 times)...(1)) >= sin(..(n times)..(1) or (sin( ...(n+1 times)...(1)) = 1 (contrad.). C2: Since sine is an odd function, -sin(..(n times)..(1) = sin(..(n times)..(-1), and applying arcsines as in C1, we have sin(1) = 1, therefore (sin( ...(n+1 times)...(1))/sin(..(n times)..(1))^2 < 1 So each term in the sequence is strictly less than 1, but it is not clear that the limit is so. This analysis assumes such a limit exists and is strictly less than 1. Your analysis would suggest that this is not the case. I suppose it is plausible that the limit is 1, in such case the ratio test would be inconclusive. One could show this using an epsilon-N type limit proof.

1:25 Here the remainder is in the Lagrange form. See Wikepedia page for Taylor's theorem: en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder

0:36 we see that sin x = x

Those A3 and B3 Putnam questions are like mathematical terrorists.

Who is Putnam?

first view !