A periodically perturbed harmonic series... does it converge???

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[@AndrejPanjkov]

I think this would have been clearer with an overview of the proof plan at the beginning.

[@samueldevulder]

So it diverges because when n gets bigger there are a whole lot of sin(n) which are lower than 1/n (in absolute value), and that makes the series diverges. Is that correct ?

[@hybmnzz2658]

Yes. We are looking at the first 2^n numbers on the number line, and showing that a decent proportion (|A_n|>2^n / 7n) of those numbers satisfy |sinx|

[@avdelft]

@@hybmnzz2658 TY. With your remark and Samuel's, Michael's video would have been be a bit clearer.

[@AndriiGudyma]

ok, this is more complicated than other videos.

[@Kapomafioso]

Before watching, I'd say it diverges, intuitively; there are many values of |sin(n)| that are "close enough" to zero (|sin(n)| ~< 1/n) to make the exponent not big enough. Since there's enough of such values, they eventually contribute enough for the sum to diverge. A formal proof is probably pretty wild.

[@aurinkona]

definitely thought it would converge before clicking. once i saw that \sum 1/(n^(1+1/n)) diverges, though, it was clear. A={|sin(n}| | n\in\NN} is dense in [0,1], which means there is an abitrarily large number of elements in A that are arbitrarily close to 0. means it's bigger than the above sum that diverges. this would suffice as a proof in most textbooks, but it's great that these expositions are available.

[@drdca8263]

Eh? But if you have some f(x) which takes values in [0,1], and such that {f(n) | n is a natural number} is dense in [0,1], but such that the closer epsilon is to 0, how infrequently f(n)< epsilon, quickly becomes more and more infrequent.., I would think that in that case, sum of n^{-(1+f(n))} would converge. So, I don’t think the density of the image of the integers under sine, is sufficient to justify the conclusion.

[@jursamaj]

@@drdca8263 It doesn't matter how infrequently f(x)

[@drdca8263]

@@jursamaj for any positive natural density, sure, but if the natural density of {n s.t. f(n)

[@lupifa4395]

A3 does not contain 4 and 5, I checked, it's above 1/3.

[@krisbrandenberger544]

Yes. Therefore, B_3 should be the singleton 6.

[@frankjohnson123]

Cool and a bit intense, nice! Would be interested to see a similar sum where the convergence depends on the value of a certain parameter, kind of like a Riemann zeta function.

[@mtaur4113]

My first thought was Ergodic theorem. Basically, long-term time average of f sampled in a well-distributed way is the same as the space average over the domain of f, given that strong enough regularity conditions hold. I don't know it well enough to do it rigorously from memory, but it probably checks out. So take this as heuristic with some steps missing. The idea is that sum probably converges if and only if the integral_[0,1] sum_{x=1,infty} x^(-1-t) dt converges. (The integral [0,1] dt is taking a rough average of |sin(x)|, in the sense of convergence testing only). The main thing is that exponent uniformly and consistently samples near 0 over the long term. To finish the argument, the x-sum can also be replaced by an integral dx, by standard comparison test stuff, and the dx dt integral diverges by direct computation. Missing some details, but this double integral diverges as expected, which is also true of the single sum_{x=1,infty} x^{1+1/x}.

[@gregdeboer1]

There's some overlap between 3 and 4 minutes when defining A2 and B2

[@yuseifudo6075]

سمعت غروبي

[@nahuelcaruso]

It's quite nice this problem, and to be honest, it was a bit challenging at first glance. But, I think that in the final part when Michael split the sum into a sum of sums (15:51) there is a kind of misleading about the index since the relationship |An|>=(2^n)/(7n) holds for n>=7, but for indexes in the sum is valid for n>128. It doesn't change the result, but it kind of rearranging must to be done carefully.

[@digxx]

When you do the re-indexing at the end, you actually retain a term |A_N|/(2^{N+2}), which you did not write, but this is finite in the limit, since |A_N|

[@Alex_Deam]

He points out that you can throw away that term because it's positive and it's an inequality

[@nodrogj1]

I think you should have drawn the unit circle arcs in your example much smaller, since in the cases you're looking at with n>7, each would only be a few degrees. I get that it's arbitrary to an extent, but if you're going to use a visual tool, it should probably aid comprehension. The fact that you chose an arc close to 1/7th the circle made it less clear that these arcs were actually meant to behave more like 1/n than 1/7, and I had to actively ignore the visual before i got what the math was doing.

[@Happy_Abe]

@19:00 why is it 2^(n+1)?

[@frankjohnson123]

A few steps aren't shown, but basically you substitute 1/k with 1/2^(n - 1) and 1/k^(|sin(k)|) with 1/k^(1/n) and then [1/2^(n - 1)]^(1/n). After some slightly messy manipulation of the exponents you get that the product is 1/2^(n + 1) multiplied by a factor greater than 1 that we can drop because of the inequality.

[@Happy_Abe]

@@frankjohnson123 wonder what those messy calculations were, I was trying to calculate

[@davidchung1697]

| sin n | is a giveaway that the serries will track the harmonic series and hence diverges. If the series had sin n, instead of | sin n |, the resulting series would still diverge.

[@jazzjohn2]

If you assume the argument n of the sine function is in degrees, then sin(n) will be zero for countably many times. Therefore a version of the harmonic series is a subset of the all positive series so it diverges.

[@MCMCFan1]

so beautiful

[@ignaciohenriquez9513]

I don't think this is true, for example it could happend that the version of the harmonic series which you end up with is the geometric series sum_{n=0}^\infty 2^{-n}.

[@jazzjohn2]

@ignaciohenriquez9513 the terms of the series would be 1/(180n) which is the harmonic series divided by 180.

[@Ligatmarping]

@@ignaciohenriquez9513 Yeah, but just in this case it diverges, since it's every 180 n.

[@sstadnicki]

You actually don't have the freedom to make this assumption mathematically - that is, mathematically (especially in analysis) we effectively take the degree sign (if it shows up) as a symbol meaning 'multiply this value by pi/180'. So in sin(90°) what we're actually doing is taking the sine of 90*(pi/180) = pi/2. This matters because many of the most important equations involving the sine function implicitly assume a 'dimensionless' (i.e., radians) argument. For instance, the approximation sin(x)≈x for small x, as well as the derivative sin'(x) = cos(x), both require this.

[@juanmanuelmunozhernandez7032]

What about replacing the exponent with 1+pi(n), where pi( ) is the prime counting function? Does it converge then? This will lead to a proof of the Riemann Hypothesis.

[@tahirimathscienceonlinetea4273]

Hi micheal this problem is really a killer problem 😊

[@567secret]

Makes me curious for what epsilon 1/n^(1+epsilon+sin(n))

[@shohamsen8986]

Awesome problem with a really amazing solution.

[@CM63_France]

Hi, Ok, but there is another series similar to that one that, I think, converges : take the alternating harmonic series, replace the factor (-1)^n by (e^(i pi))^n , and then replace e^(i pi) by any unit complex number. I think that serie converges, obviously except for this coefficient = 1 . For example if this unit complex coefficient is equal to i, you get pi/4 + i ln 2 , some thing like that. And may be it is not necessary for this coefficient to be a rational power of i, or -1.

[@Noct1um]

Yup, you can calculate the analytic continuation for your series in the variable z, that you called unit complex number(try to find the derivative in z). And indeed the series converges exactly for |z| less or equal to 1, except when z=1.

[@CM63_France]

@@Noct1um May be Michael will make a video about that.

[@Jack_Callcott_AU]

Nice❕ I must watch it again to understand better. 🧑‍🔬

[@DrR0BERT]

There's a glitch in the matrix at 3:37.

[@artsmith1347]

I would not have expected this to diverge: sum from 1 to infinity of (1/ n^(1+1/n)) But I found a webpage that demonstrates it. I would post a link, but YT doesn't like it when I post links. Never mind that the internet is about sharing information -- but only if it is "approved" by the Ministry of Truth.

[@tomkerruish2982]

Your thoughtcrime has been reported to the Ministry of Love. Resistance is futile.

[@matchamitminze]

It’s just because KZbin channels (by default) prevent comments with links in them from being posted, since there’s been a long history of phishing and scam attempts on bigger KZbin channels in the past, usually in the form of “giveaways.” It’s not some conspiracy, and it takes a very small amount of time and going out of your way to learn.

[@Ligatmarping]

It's relatively straightforward from the comparision criteria. The general term divided by 1/n is nth root of n which goes to 1 as n goes to infinity. Therefore the series behaves as the harmonic series 1/n, which diverges.

[@pavlopanasiuk7297]

But it was demonstrated on this channel already

[@mathisnotforthefaintofheart]

This can be shown very easy with limit comparison test with the Harmonic series. Only few lines and done

[@ThePayner11]

Hi, I need help with this problem, and I'm not sure how to solve it...I made it up myself, and it was quite interesting attempting it. Find the area bounded by arccos⁡(e^x), arcsin⁡(e^x) and the y-axis.

[@BederikStorm]

if you try the sum of n^(-1-|sin(n)|), it will converge.

[@Holasiquetal]

Please: a video about the convergence of the sum(p=2 to infty) 1/p^s. Where p are the primes. For which value of s are we sure that the sum converges? It looks to be between 1.3 and 1.5. This value looks interesting.

[@ronald3836]

This sum diverges for s 1, so just like sum_n 1/n^s. A recent video showed divergence for s=1. The quick proof is to take Euler's identity sum_n 1/n^s = prod_p 1/(1-p^(-s)). Now take the log of both sides. The LHS converges for s > 1 and diverges if you let s go down to 1, so same for the RHS. The RHS is log(prod_p 1/(1-p^(-s))) = sum_p -log(1-p^(-s)), and now use -log(1-x) = x + x^2/2 + x^3/3 + ...

[@ronald3836]

"so same for the RHS" is a bit tricky, since interchanging limit and infinite sum is not always allowed. I am pretty sure it can be justified here.

[@martinepstein9826]

If s > 1 then the sum converges since it's less than sum(n = 1to infty) 1/n^s. Michael briefly explains why this converges at 0:23 If s = 1 then the sum diverges. See Michaels video "A harmonic series with only primes".

[@Holasiquetal]

@@martinepstein9826 From France, thank you all 👌

[@China-Voice]

A3 & B3 of course contains "7"

[@Alan-zf2tt]

Strange noise @ 18:19

[@TheEternalVortex42]

The sum of 1/n^(1+1/n) diverges as noted in the beginning, but it seems "clear" that sin(n) is going to be much larger than 1/n for large n, so it seems kinda obvious that this one would diverge as well. (Indeed the proof in the video makes rigorous that inequality).

[@randomclips9715]

wouldn’t a larger exponent mean it’s more likely to converge

[@Nebula_ya]

Yeah, sin(n) being larger makes it more likely to converge

[@woody442]

This Argumentation is flawed anyways, because sin(n) is periodic with 2 pi, abs(sin(n)) periodic with pi. It might be that transcendency of pi screws your loose assumption of sin(n) being larger than 1/n for integer n's.

[@bjornfeuerbacher5514]

"but it seems "clear" that sin(n) is going to be much larger than 1/n for large n" This isn't cleat at all - and in fact, it's wrong. Where did you get that idea from?

[@ronald3836]

@@bjornfeuerbacher5514 and if true, it would lead to the opposite conclusion.

[@minhnguyen1338]

can you prove that 11162 problem

[@atzuras]

I am amused by geometric proofs. But real life perturbations are more like infinite colours of noise riding a stallion on drugs.

[@monke9865]

What ?

[@botsjeh]

a glitch in the matrix

[@charleyhoward4594]

Tour de Force

[@pierreabbat6157]

Your 2π/7 looks like 2π/9.

[@Double_U_tau_Phi]

First

[@primenumberbuster404]

First

[@drdca8263]

fIrSt!!1!!1!