Does this infinite series converge? (I solved it!)

Рет қаралды 12,571

Күн бұрын

We will look at a particular infinite series involving the sine function. We will determine whether the series converges or diverges. The solution involves rational approximations of the number π.
Having some knowledge of trigonometry and infinite series will be useful while watching this video.
Timestamps
00:00 Introduction
08:25 Tame Numbers versus Wild Numbers
11:34 Controlling the Wild Numbers
19:18 Mahler's Theorem
23:59 Conclusion
Correction: 00:40 The digits of pi are written incorrectly. The sequence 2643383279 was inadvertently repeated.
Ravi B. Boppana (2020), "Convergence of a sinusoidal infinite series from Borwein, Bailey, and Girgensohn", arxiv.org/abs/2007.11017.
Ravi B. Boppana (2005). Posts by user Ravi B on the thread “Tough infinite series” on the Art of Problem Solving forum.
artofproblemsolving.com/commu....
Kurt Mahler (1953), "On the approximation of π", Indagationes Mathematicae (Proceedings), 56, 30-42.
Gregory V. Chudnovsky (1982), "Hermite-Pade approximations to exponential functions and elementary estimates of the measure of irrationality of π", in The Riemann Problem, Complete Integrability and Arithmetic Applications, pages 299-322. Lecture Notes in Mathematics 925.
Doron Zeilberger and Wadim Zudilin (2020), "The irrationality measure of π is at most 7.103205334137 . . . ", arxiv.org/abs/1912.06345.
Jonathan M. Borwein, David H. Bailey, and Roland Girgensohn (2004), Experimentation in Mathematics: Computational Paths to Discovery. CRC Press.
Wikipedia article on Root Test and other standard "Convergence Tests":
en.wikipedia.org/wiki/Converg...
Photo of Kurt Mahler from 1970 due to Konrad Jacobs.
To stay up to date, subscribe to this KZfaq channel. Thanks!

Пікірлер: 128

@mohannad_139 2 ай бұрын

The only thing that's wild is the fact that this only has 1,000 views

@diegoman8158 2 ай бұрын

This is how real problem solving feels when doing research. Thank you for remembering me that feeling, that was a beautiful solution!

@matthewsarsam8920 2 ай бұрын

It’s 3am and I just learned this series converges

@marchrushart6594 2 ай бұрын

Extremely underrated channel hope this video get on KZfaq's algorithm radar

@MathTutor1 2 ай бұрын

Congratulation on a great video. I'm the 165th subscriber. At this time of seeing the video, there are 5236 views to your video. This is a great number too. The first is the sum of the next two and the last is the product of the previous too. You will get a lot more subscribers soon because your explanations are great. Thank you.

@graf_paper 2 ай бұрын

This was such a nice presentation. Very well paced, and the explanations were really easy to follow while being just interesting enough that I kept stopping the video to ask what other problems your method might solve.

@BoppanaMath 2 ай бұрын

Glad you enjoyed it!

@BoppanaMath 2 ай бұрын

If you're looking for other problems, here is a generalization that seems to be unsolved: mathoverflow.net/questions/409550/convergence-of-the-series-sum-n-1-infty-frac2-sin-nn3n-na-for

@gabrielafrajtag86 2 ай бұрын

This was so cool to watch!! Thank you!!

@sankasr 2 ай бұрын

Incredible video! Please keep it up Sir, maths content is always needed.

@abhaygvasista9433 2 ай бұрын

man, ur channel is underrated

@user-bk3gy9sw4b 2 ай бұрын

I was trying to solve before I watched the video. I had tried every inequality I know to get rid of the annoying sine function but they all ended up with the harmonic series, which, as you already know, does not prove anything. I got nowhere to go, so I started wandering if there is anything to do with the distribution of numbers at which sin value is close to one. Eventually I gave up and after I watched the video, it turns out I was kind of on the right track! It really surprises me. Now I know it would be helpful to consider how frequent the out-of-control numbers pop up. Your great work inspires me.

@Nekathe 2 ай бұрын

It was 1:30 am and I was trying to sleep… I found this video and now it’s almost 2 am… great video man

@inheritance1097 13 күн бұрын

Keep posting man. Your videos are great.

@BoppanaMath 13 күн бұрын

Thanks! Yes, I will keep posting. My next video is taking longer than expected, but I hope to put out a video every month or two.

@ImMataza 2 ай бұрын

Very interesting video. Thank you for making it!!!

@quantum1861 2 ай бұрын

Fantastic overview!

@qcard76 2 ай бұрын

Very cool. Your math succeeded in proving the series’ convergence where the likes of Wolfram Alpha thinks it diverges

@mrguidetolife 2 ай бұрын

At 18:30 you conclude that 'gap' must be of the form ±8/∜(n)+2kπ, but this isn't necessarily true. Call the two wild numbers 'n' and 'm' instead, such that m=n+d, where 'd' is the gap between them. Then by your previous proof it follows that n=½π±/∜(n)+2kπ and m=½π±/∜(m)+2pπ. Since d=m-n it follows that d=±4/∜(m)∓4/∜(n)+2xπ. This might not cause problems in the proof but the distinction is important none the less.

@BoppanaMath 2 ай бұрын

You are right to be cautious, but it's all good. First, my notation ±epsilon means "at most epsilon in absolute value" (as opposed to "exactly equal to epsilon in absolute value"). Second, because m > n, any number that is ±4/∜(m) is also ±4/∜(n). Third, any number that is ±4/∜(n) - ±4/∜(n) is also ±8/∜(n). Make sense? I was more careful with the notation in my paper: arxiv.org/abs/2007.11017. See the bottom of page 3.

@bozzo.smalls 2 ай бұрын

Excellent video

@rpggamers7867 2 ай бұрын

This video is so inspiring.hope I'd reach such levels if mathematical mastery

@carstenmeyer7786 2 ай бұрын

20:02 That approximation looked oddly similar to a criterion for algebraic numbers from continued fractions theory... looked it up, and yes, in the (German) lecture "Kettenbrüche" by Tomas Sauer (2005) the following is proven on p.34: -------------------------------------- *Theorem 2.32 (Liouville):* Let *x* be a real algebraic number of order *n.* Then there exists "C > 0" such that *|x - p / q| > C / q^n for all p in Z, q in N* -------------------------------------- This theorem shows algebraic numbers cannot have arbitrarily fast converging continued fractions expansions. It's interesting "Mahler's Theorem" shows the converse is not true [for pi] -- pi is transcendental, but still has a continued fraction converging as slowly as algebraic numbers of order 20.

@BoppanaMath 2 ай бұрын

Yes, great points. Roth actually improved Liouville's theorem to show an exponent of 2 + epsilon: en.wikipedia.org/wiki/Roth%27s_theorem. For more about the "irrationality measure" of specific numbers, see en.wikipedia.org/wiki/Liouville_number#Irrationality_measure .

@carstenmeyer7786 2 ай бұрын

@@BoppanaMath Amazing, I learned something new, thank you for providing a more accessible reference!

@user-ei6rd7ei7x 2 ай бұрын

The tame-wild idea is brilliant! Does the series ((1-eps)+eps sin(n))^n / n converge for all eps > 0?

@BoppanaMath 2 ай бұрын

Good question! Yes, for every 0 < eps

@user-gj2kw7wx2q 2 ай бұрын

Honestly you're sick in presentation of the solution, I was engaged the entire length of the video:D

@Joffrerap 2 ай бұрын

very nice. You've made some complicated math very accessible, for a problem that anyone can understand. What if you put a sqrt(n) in the denominator? or how low could we get for an exponent alpha in the denominator while still having a convergent series? My intuition is that (2/3+1/3*sin(n)) will not get close to 1 often enough to compensate for putting to the power n, so that with alpha = 0, it would converge as well.

@BoppanaMath 2 ай бұрын

Thanks and good questions. For alpha >= 0.92, I can show convergence using the same method. If the irrationality measure of pi were 2 (an open question), then alpha > 0.5 is enough to show convergence. I'm pretty sure that alpha = 0 gives a divergent series, with the terms not even heading to 0. For example, when n = 51819, the numerator (including the power of n) is bigger than 0.99999, quite far from 0. Here is a link to another person who asked your question: mathoverflow.net/questions/409550/convergence-of-the-series-sum-n-1-infty-frac2-sin-nn3n-na-for.

@animowany111 2 ай бұрын

This video popped up in my recommended a couple times, I skipped it earlier because I thought it was one of those math channels solving simple problems on a whiteboard, but now I actually tried to think about it, "Surely, the answer is obviously... wait, it's not obvious". For now I'm adding this to my watch later list while I stew over the problem for a bit, I wonder if you can use the continued fraction approximations for pi to see if the close multiples keep giving large series terms as n grows. By itself that could be enough to prove non-convergence if things go well. Proving convergence would be harder unless there's a way to group things up nicely.

@BoppanaMath 2 ай бұрын

Without giving away the answer, I will note that the numerator seems to be large infinitely often. For example, when n = 51819, the numerator is bigger than 0.99999. As you guessed, I used the continued fraction for pi/2 to come up with 51819.

@animowany111 2 ай бұрын

@@BoppanaMath Oh, I completely forgot about the division by `n` when thinking about nonconvergence. This does make me think it's highly likely this series converges, but proving that remains non-obvious. At the time I wrote that comment I also forgot the simple continued fraction for pi is not nice, but I think my intuition for how to approach convergence still holds. I think you can show an upper bound on the series by summing multiples of the pi/2 convergents (I think showing each one is upper bounded by a certain geometric series should suffice), and showing the junk in between contributes ~nothing.

@user-vg1qo5gi3l 2 ай бұрын

Congratulations! I'm a little bit surprised that you posted a solution only after 15+ years

@BoppanaMath 2 ай бұрын

Thanks! Yes, when I solved the problem in 2005, I was taking a break from academia, and so wasn't focused on writing papers. Fun fact: when I first solved the problem, I didn't know it was an open problem. Only a year later did I learn that it was posed as an open problem in a book.

@jmarvins 2 ай бұрын

@@BoppanaMath how'd you find the problem if not through the Borwein et al book? great video, in any case!

@BoppanaMath 2 ай бұрын

@@jmarvins The problem was posted on the Art of Problem Solving (AoPS) forum in December 2004, without any indication that the problem was open or from a book. Maybe it was better that I didn't know the problem was unsolved, since otherwise I might have been intimidated, ha ha. Here is the AoPS link: artofproblemsolving.com/community/c7h22093.

@user-vg1qo5gi3l 2 ай бұрын

@BoppanaMath Do you think that your approach may help to solve Flint Hill series?

@BoppanaMath 2 ай бұрын

@@user-vg1qo5gi3l I don't know. Zudilin and Alekseyev have shown that whether the Flint Hill series converges is related to the irrationality measure of pi (which I also needed). mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1 arxiv.org/abs/1104.5100, In particular, they showed that if the Flint Hill series converges, then the irrationality measure of pi is at most 2.5 (which would be a dramatic improvement of known bounds). It's possible that my analysis of gaps could show the converse of that implication. If interested, you should give it a try!

@dimitrilemeur7703 2 ай бұрын

Thank you so much for the video. I have a somehow related question. For which alpha the sequence (n^alpha sin(n)) admits a convergent subsequence ? When it is not the case, how fast grows the sequence a_n = inf | k^alpha sin(k) |, where the inf is taken for k greater or equal than n ? For instance what happens in the case alpha = 1 ?

@BoppanaMath 2 ай бұрын

For 0

@marcusrosenberg3255 Ай бұрын

I love your videos.

@BoppanaMath Ай бұрын

Thanks! Next video coming in a few days.

@petrie911 2 ай бұрын

A natural extension of the problem is to replace sin(n) with sin(2pi n / T), for some irrational period T. This shows that if T, has finite irrationality measure, then the series converges. But what if T is a Liouville number? Does the series still converge?

@BoppanaMath 2 ай бұрын

Great question. I'm guessing there are Liouville numbers for which the series diverges. But I don't know for sure.

@djridoo 2 ай бұрын

Waw, amazing !

@tomdeneckere 2 ай бұрын

18:23 Shouldn’t the second equation have n+gap in the fourth root instead of just n? Or doesn’t that matter?

@BoppanaMath 2 ай бұрын

Good point. Yes, you are correct. Fortunately, 4 over the fourth root of n+gap is less than 4 over the fourth root of n, so the error bound in the video is also correct. I chose not to mention this subtlety in the video, but it is mentioned in the paper. You are observant!

@markmajkowski9545 2 ай бұрын

Isn’t 2/3 + 1/3 sin n

@BoppanaMath 2 ай бұрын

You're right that 2/3 + 1/3 sin n is always less than 1. But it can get as close as we want to 1. There is no p < 1 such that 2/3 + 1/3 sin n is always at most p. So we can't compare to a specific geometric series.

@bozzo.smalls 2 ай бұрын

What software did you use to make such a clean presentation?

@BoppanaMath 2 ай бұрын

Glad you liked the presentation! I used the Beamer class of LaTeX to create the slides. I used Asymptote to create the graphics.

@KrasBadan 2 ай бұрын

Very cool

@denki2558 2 ай бұрын

By intuition, it converges. Upper bound is the harmonic series if sin n is always 1, but sin n will never equal 1 since n is an integer. So as n grows large, the graph of the numerator will look like sums of dirac deltas every 2pi but those will never hit integer inputs. Though even I'm not convinced by this because I ignored the denominator. Edit: Great vid!

@BoppanaMath 2 ай бұрын

You have the right intuition. But as your last sentence says, you have to be careful, because without the denominator, the series would diverge.

@user-hi9zx2dw6s 2 ай бұрын

A little question, since you use Mahler theorem in your proof, how do you know the gap is an integer ?

@BoppanaMath 2 ай бұрын

I'm a little puzzled by your question. The gap, by definition, is the difference between two consecutive wild numbers. Wild numbers, by definition, are integers. So the gap must be an integer too.

@user-hi9zx2dw6s 2 ай бұрын

⁠⁠⁠⁠@@BoppanaMath Oh, thanks. I’d stuck on 17:36 part, the equality made me think that n would be irrational. Really appreciate your response.

@BoppanaMath 2 ай бұрын

@@user-hi9zx2dw6s Got it. When I write ±something, I mean "at most something in absolute value" (as opposed to "exactly equal to something in absolute value").

@Asiago9 2 ай бұрын

I might be getting the math wrong on this, but I used the limit comparison test having a_n as the ((2/3+1/3sin(n))^n)/n, and b_n equal to the alternating harmonic series, and dividing a_n by b_n getting ((2+sin(n))^n)/(-3)^n which as n tends to infinity will get closer to 0, as the most the top will be is 3, and then the absolute value of the bottom is always equal to or bigger than 3, and is alternating, therefore since the sum of (-1)^n/n converges due to the alternating series test, ((2/3+1/3sin(n))^n)/n, converges due to the limit comparison test

@BoppanaMath 2 ай бұрын

Doesn't the limit comparison test require that both series be nonnegative? en.wikipedia.org/wiki/Limit_comparison_test. I also disagree with your claim that ((2+sin(n))^n)/(-3)^n heads to 0. For example, when n = 51819, this expression is about -0.999995, quite far from 0.

@GreenMeansGOF 2 ай бұрын

I’m guessing that none if the standard approaches work? How about the extensions of the ratio/root test? Is there a more elementary proof?

@BoppanaMath 2 ай бұрын

As far as I know, none of the standard approaches work. As the beginning of the video shows, the root test is inconclusive. Since the root test is inconclusive, the ratio test is inconclusive too. Convergence of the series is essentially equivalent to showing that pi is not too close to rational numbers. At a minimum, that involves showing that pi is irrational. So the proof can't be too elementary. I'm not aware of a more elementary approach.

@regulus2033 2 ай бұрын

Well, it's easy to say "There is a theorem, so using it, we can show that..." But can we prove it from scratch, using only pen and paper?

@BoppanaMath 2 ай бұрын

Are you referring to Mahler's theorem? I listed Mahler's original paper and two improvements in the video description. For example, here is the paper by Zeilberger and Zudilin: arxiv.org/abs/1912.06345. It's a 13-page paper with all details of the proof.

@regulus2033 2 ай бұрын

@@BoppanaMath Thank you for the answer! I mean, usually to determine convergence of a series, you don't need to know an irrationality measure of pi. Is there another way?

@BoppanaMath 2 ай бұрын

@@regulus2033 You're welcome. I don't know of another way. Somehow, we need to show that sin(n) is rarely too close to 1. That's basically the same as showing that pi is rarely near a rational number with numerator n, so irrationality measure seems relevant.

@Czeckie 2 ай бұрын

pretty cool

@mcpecommander5327 2 ай бұрын

Haven’t watched yet. sin(integer) is always

@BoppanaMath 2 ай бұрын

The angles are in radians (as the video mentions near the start). The numerator is less than 1, but that's not enough to deduce convergence, because the numerator gets arbitrarily close to 1. For example, when n = 51819, the numerator (including the exponent) is bigger than 0.99999.

@mcpecommander5327 2 ай бұрын

I see! Indeed, the proof in the video is quite insightful

@datfry7791 2 ай бұрын

beautiful proof! but i don't get why you can't just say that since there's no integer equal to any multiple of π/2 (i think), sin n is never equal to 1 for any integer, which gives us that 2/3 + (sin n)/3 is in the radius of convergence of the power series Σ(x^n)/n?

@BoppanaMath 2 ай бұрын

You are right that sin n is less than 1, and that 2/3 + (sin n)/3 is less than 1, but both can get as close as we want to 1. For example, when n = 51819, both are bigger than 0.999999999. To get your argument to work, you would need to find a positive number epsilon such that sin n is always less than 1 - epsilon, but there is no such epsilon.

@jellyfrancis 2 ай бұрын

Since n is an integer it can never be a odd multiple of pi/2 because its transcendental...by that the numerator can never be 1 ...so the root test wroks ... Can't we say that ?❤

@BoppanaMath 2 ай бұрын

The numerator is less than 1, but it can be arbitrarily close to 1, so the root test doesn't work.

@theultimatereductionist7592 2 ай бұрын

Kurt Mahler made prominent contributions in defining classes of numbers: E-numbers, M-numbers, based on varying levels of transcendentality.

@vekyll 13 күн бұрын

I've read the paper and it is correct there, but in the video some things are too simplified and not correct in pure sense. The error starts at 17:30, of course n is not _equal_ to pi half plusminus four over qroot n plus period, it is _between_ those points. And of course, by 18:30, the error is quite obvious--the consecutive wild numbers can even enter the same interval around a concrete translate of pi half (as happens at the begining, with 2&3;7&8&9;...), let alone one being near the end of the interval while other being near the beginning.

@BoppanaMath 13 күн бұрын

Fair points. My notation ±epsilon means "at most epsilon in absolute value" (as opposed to "exactly equal to"), like how it is used in statistics and experimental science (but not how it is used in say the quadratic formula). I say it at 17:36, but maybe I should have written an explanation on screen. About the initial wild numbers (2, 3, 7, 8, 9), I avoid them with my temporary assumption that n > 5000 at 18:04. I remove the assumption at 21:40. I agree that the paper is more formal.

@VeteranVandal Күн бұрын

Wild result.

@Almondz_ 2 ай бұрын

Do you have a discord?

@BoppanaMath 2 ай бұрын

Sorry, I don't.

@prince_gamer580 Ай бұрын

At 0:41 the value of pi written is wrong, after 43383279 it should be 50288 whereas 2643383279 is repeated in between

@BoppanaMath Ай бұрын

Oops! You are correct. I inadvertently repeated the sequence 2643383279. I just included a correction in the video description. Thanks for your eagle eyes!

@prince_gamer580 Ай бұрын

@@BoppanaMath happy to help 😄

@olegmazonka1869 2 ай бұрын

Converges to 2.1631961

@BoppanaMath 2 ай бұрын

Thanks. Your calculation is consistent with the bottom of the following MathWorld page, which says that the sum of the first 10 million terms is approximately 2.163: mathworld.wolfram.com/HarmonicSeries.html. How many terms did you calculate?

@olegmazonka1869 2 ай бұрын

@@BoppanaMath I did not calculate terms directly. I represented Pi as an approximate ratio and rearranged the sum to go by the list of the selected angles then the infinite sums of Harmonic-Geometric series that are the same as Hurwitz functions. Those can be computed quite precise without going very far in terms. Then one can see how the result converges while having Pi approximation more and more precise. There is even better way of computing this sum by slicing the circle in halves with smaller slices converging to Pi/2 point. Then n can be computed for each circle point as a reciprocal in modular arithmetic, where the modulus is the precision and the numerator in the approximation of Pi. However, I did not implement this method. I am not sure if this all makes much sense without detailed explanation.

@BoppanaMath 2 ай бұрын

@@olegmazonka1869 Thanks for the explanation! Sounds intriguing. Does your method give an alternative proof of convergence of the original series?

@olegmazonka1869 Ай бұрын

@@BoppanaMath I just saw this video kzfaq.info/get/bejne/iMWFmZpp2M-cmp8.html and thought maybe it is possible to make binomial expansion and then compute powers of sine independently.

@ArthurvanH0udt 7 күн бұрын

Around the 16m20 to 16m30 point. I know/see the cos(E) thing but you go a bit too fast there imho. Then at 16m30 the divided by 5 imho is too unexplained!

@BoppanaMath 7 күн бұрын

Thanks for the feedback. Take a look at my paper for more details: arxiv.org/abs/2007.11017. In the proof of Lemma 1, I show that cos(theta) is at most 1 - (2/pi^2) theta^2, which is at most 1 - theta^2/5 (for theta between -pi and pi).

@ArthurvanH0udt 7 күн бұрын

@@BoppanaMath yup, I already found that (based on your title info). THANKS!

@PRABALBAISHYA-xi1fd 2 ай бұрын

A non rigorous way to conclude that the series converges is to observe that the numerator term will always be less than 1. This is so as the maximum value of the numerator can be at most 1 when sin(n)=1, but for this to happen n must be of the form (4k+1)*π/2. But since we are summing over the natural numbers n will never be of the this form. Hence after concluding that the numerator will forever stay

@BoppanaMath 2 ай бұрын

Unfortunately, this approach doesn't work. You're right that the numerator is less than 1, but it repeatedly stays large. For example, when n = 999,826, sin(n) is bigger than 0.999999. At the beginning of the video, I showed that the root test is inconclusive here. It's known that if the root test is inconclusive, then the ratio test is inconclusive too,

@MH-sf6jz 2 ай бұрын

Consider the sequence 1/(1+1/n) * 1/n, this satisfies your “forever < 1” claim, but this is clearly divergent sum because 1/(1+1/n)>=1/2

@kizyzo1348 2 ай бұрын

@@BoppanaMath I didn't explicitly use the root test here. But still at n = 999826 sin n may be really close to one but the whole numerator is equal to 0.89227. Something that I' ve noticed is that when we multiply the terms of the harmonic series with some 'weight', with |weight|

@kizyzo1348 2 ай бұрын

@@MH-sf6jzumm I don't think it satisfies my claim that the numerator is always less than 1 since it is exactly equal to 1. Moreover, as n gets large we are simply summing a bunch of 1s so the series is clearly divergent

@BoppanaMath 2 ай бұрын

@@kizyzo1348 You seem to be claiming that when weights are less than 1 (in absolute value), then the weighted harmonic series converges. But that's not true. For example, all the weights could be 1/2.

@deltalima6703 2 ай бұрын

Yes, its trivial. You can just look at it and know.

@BoppanaMath 2 ай бұрын

Got it. Maybe you could solve the Flint Hill series, which is currently an open problem. arxiv.org/abs/1104.5100.

@KipIngram 2 ай бұрын

I don't see how it could not. The sin() function will only reach it's maximum at pi/2, 3*pi/2, 5*pi/2, etc. - the argument to the sin() is an integer, so it will never hit any of those. So that term will never, ever contribute the full 1/3. Therefore, the quantity in parentheses will always be in the range 1/3 < x < 1, without equality occurring ever. So it will always be less than one. Then we are raising it to an ever growing power and dividing it by an ever growing denominator. It has to become small. I realize this argument is in no way rigorous, but then again I am not a mathematician. I'm an engineer. And if I were faced with this I would proceed on the basis that yes, it converges.

@BoppanaMath 2 ай бұрын

Yeah, that's a fine intuition. But consider n = 51819. Its sine is bigger than 0.999999999. Even after you raise to the power n, the numerator is bigger than 0.99999. How do we know that such cases don't happen often? That's why we need a rigorous argument.

@deltalima6703 2 ай бұрын

Divided it by 52000 or so. Its still teeny. I dont see the potential either.

@deltalima6703 2 ай бұрын

Yup, 2 engineers solved it in under a minute. Thats why bridges dont fall down unless you hit them with container ships or something. Note: just throwing shade, video was interesting. I know about what a windy day can do, obviously.

@cowlikenuts 2 ай бұрын

This is a fantastic video. I almost felt like I was at a seminar at a university's pure maths department, which is to say that this video was presented very clearly, professionally and engagingly (although the latter adjective doesn't always hold, for said seminars). The idea of turning an mathematical concept that a first-year student may grasp, such as convergence of an infinite sum, and then using advanced machinery, such as Mahler's theorem, was genuinely surprising to see. I hope you continue to make interesting maths videos, and that similar videos get a lot more attention.

@BoppanaMath 2 ай бұрын

Thank you so much for your kind words! Yes, I am planning to create one video a month.

@1.4142 2 ай бұрын

18:25 voice crack

@mcumer 2 ай бұрын

A non- rigorous proof: the average value of sin x is zero.. therefore we can subsititute sin n with zero ( the infinite values of sin n have random values)

@BoppanaMath 2 ай бұрын

Got it. Now consider the same series without the denominator of n. Your heuristic would suggest that the modified series also converges. But the modified series diverges.

@Nebukanezzer 2 ай бұрын

I think it's quite easy to see that this converges. (Haven't watched yet, may be wrong.) The harmonic series is near the boundary of what converges or diverges. Make the top grow more slowly or the bottom grow more rapidly and it stops diverging. (Sort of. 1/(n log n) diverges but 1/n^1.0001 converges.) Now, the interior of the numerator will never be exactly 1, as there is no integer N such that sin(N) is 1, so the interior of the numerator is strictly less than 1. It is also exponentiated, and that's kinda sorta like the geometric series, which converges. And if the numerator converges, the obviously the whole thing is going to. Now, the interior of the numerator is going to brush up very close to 1 infinitely often. So it doesn't converge like the geometric series, but it definitely is smaller than if the numerator was a constant. It's kind of a race condition. How dense among the integers are the values for which sin(n) is closer to 1 than the amount by which the power of n on the outside of the numerator decreases it? If this happens quite often, then it might diverge like 1(n log n) does. But if it doesnt, then it might converge like 1/n^2. For whatever intuitive reason, I would think these values that keep the numerator large are less common than would be necessary. It comes down to the integer approximations get close enough, quickly enough, that the factor of n stops mattering and you get something like the harmonic series, or if the values that work are sporadic, and you get something like the reciprocal of the squares, and I feel like it's probably closer to the latter. Yes, I know precisely none of what I just wrote is actual math, but this is just what went through my head in the first 30 seconds of staring at the problem.

@BoppanaMath 2 ай бұрын

Your intuition is spot on. The hard part of the proof was to show that the big terms are sporadic. Sporadic enough that the big terms could be bounded above by 1/k^1.01, the sum of which converges.

@randyzeitman1354 2 ай бұрын

eh... lucky guess...

@hlee4248 2 ай бұрын

I don't think you have proved that the series converge. What you have shown is that the series is bounded, i.e. the series is approximately less than 300, but you did not show the aeries converges to a single number.

@decare696 2 ай бұрын

all the terms are positive, so the partial sums form a strictly monotonic sequence. As such, an upper bound is enough to show convergence.

@hlee4248 2 ай бұрын

@@decare696 Positive partial sums does not imply monotonic. For example, the series 1+1-1+1-1+1-1+... has positive partial sums but is not monotonic and the series does not converge.

@BoppanaMath 2 ай бұрын

@@hlee4248 Each *term* in the series is positive, so the partial sums are monotone increasing.

@hlee4248 2 ай бұрын

@@BoppanaMath & @decare696 How did I miss that! Thank you both for the patient replies.

@jocabulous 2 ай бұрын

bro sounds like dora when he ask me if i know something

@anestismoutafidis4575 2 ай бұрын

It is not usual converge because the infinite series runs against zero. It's a zero series. Σ ♾️ (n=1) [2/3 +sin(♾️) /3] ^♾️/♾️ Σ ♾️ (n=1) (2/3)^ ♾️+0÷ ♾️ =0