Every Student Should See This

This has always been one of my absolute favorite visual proofs!
This sum 1 + 1/2 +1/3 +... is call the harmonic series. This is a great math proof showing why the harmonic series diverges! (We basically use the integral test for series)
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #shorts

[@BriTheMathGuy]

Watch This Next! kzbin.info/www/bejne/qHmVqKSQap2YfNk

[@GooglyMoogly420]

The answer is 1 in this video, not infinity or 2

[@user-jf4te6kg5r]

1

[@FMRovers]

But.. Plank length..

[@nonamedpleb]

no

[@RiteshSingh-ly4wu]

Answer is infinity

[@shinobix4925]

Infinity + infinity = infinity so I guess the answer is at least 2

[@RubyPiec]

nah it's like at least 7

[@lyrimetacurl0]

It's at least TREE(G(64))

[@romi5073]

@@lyrimetacurl0 ^Rayos numbers

[@krashsytegaming9222]

@@romi5073 eulers number

[@TheMarkzon]

I feel like it's more than 1 but never 2

[@that1dudem]

Year 2, month 7, day 23: I'm starting to think this may be a loop

[@gurpreetsingh7582]

You can't be more wrong my friend

[@maxz69]

You see the red bar underneath

[@that1dudem]

@@maxz69 bruh it was a joke

[@maxz69]

@@that1dudem I know but it's not as if the bar isnt visible. It isn't a joke if it's obvious that it isn't a joke

[@bartek7383]

​@@maxz69 🤓

[@piotrosiejuk4961]

More elegant proof: 1+ 1/2+ 1/3+1/4+ (sum of those 2 is larger than 1/2) 1/5+1/6+1/7+1/8 +(sum of those 4 is larger than 1/2) ... You can keep on adding further "chains" (length 8, 16, 32 and so on), each with value of at least 1/2 (pretty obvious). This gives you an infinite number of 1/2s. This is infinity.

[@gustafbstrom]

This is better than the clip IMHO. Nice one.

[@theforbiddenfruit2300]

Yeah this made more sense than the vid

[@user-js9vh7hj4c]

Thats not a proof, thats guessing .

[@user-js9vh7hj4c]

​@@theforbiddenfruit2300the video needs atleast some calculus knowledge to make sence

[@methatis3013]

​@@user-js9vh7hj4c no, it's called Cauchy condensation criterion

[@quasarolive9136]

All infinities are great. But some infinities are greater than others. - George Orwell

[@sreekanthhere]

George Cantor

[@elchucapablas]

Not all infinites are created equal

[@quasarolive9136]

@@sreekanthhere Oh wow- actually, I was just making that stuff up after George Orwell's quote in Animal Farm "All animals are equal, but some animals are more equal than others." I didn't realize until now that George Cantor had said the same thing 150 years ago. Haha- thanks for letting me know.

[@govinddwivedi582]

One infinite good, two infinite better

[@LightningStrike691]

@@sreekanthhere “nuh-uh” - L.E.J Brouwer

[@cosinex8361]

In math, we have pretty much committed to memory: harmonic series always diverges

[@skylardeslypere9909]

I like how things like 1+2+4+8+... "make sense" when looking at the analytic continuation of the Riemann Zeta function, but even then the harmonic series still diverges. It really can't catch a break

[@elizathegamer413]

If p

[@CasanovaLucas]

CosineX yeah, but why? (the videos shows the reason about this divergence)

[@matei5929]

@@CasanovaLucas By the integral test. For a sum 1/n^p with p=1 in this case (although it applies to any value of p1, the antiderivative still has some p in the denominator so the series converges.

[@CasanovaLucas]

@@matei5929 that was a rethoric question my friend hahahaha

[@SupremeST25]

How many times are you gonna ask me bro😭😭i wanna go home

[@albert.einstein602]

fr bro is playing with us😭

[@MrDiceZack]

Infinite amount of times obviously

[@hjthegamer4445]

Bro stuck in loop 😢

[@girderzoegar8440]

😅

[@IshitaKhurana-eh8dw]

😂😂😂

[@The_Honored_One_Inf]

For the next person to say the answer is ~2, that only applies to a sequence where the bottom number doubles every time. In this sequence it goes over two right after 1/3.

[@The_Honored_One_Inf]

@seanclemson9554 supertasks are a philosophical thing, not really mathematical. I guess you could call it mathematical but usually supertasks are contradictory or paradoxical like the grim reaper supertask.

[@pineappleudh6561]

Honestly didn't believe you but yeah, it's true. 1+0.5=1.5 1.5+0.25=1.75 1.75+0.33=2.08 Does make me wonder though, when does it pass to 3? How much longer does it take to pass to the next whole number each time?

[@TurkishKS]

If you change the series by rounding each fraction down to the nearest 1/x fraction, where x is a power of 2, you'll see that this series is the same as repeatedly adding 1/2 (per the grouping in brackets I've made below). Since this clearly smaller series, can be shown to be equivalent to repeatedly adding 1/2, and thus equal to infinity, the original larger series must be infinity as well. 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 +1/8) + (1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16 + 1/16) + ...

[@sigmaoctantis5083]

That's the classical argument due to Nicolas Oresme - clearly the more elementary proof than using an improper integral.

[@babyboy5553]

I might be confused, but overall you cannot rearange terms in non-absolutely convergent series, the result will change.

[@TurkishKS]

@@babyboy5553 I rearranged nothing. I simply rounded every term down to the nearest power of two reciprocal, meaning my series is smaller. The brackets are solely to show that those powers of two can be grouped up and thought of as repeatedly adding 1/2, to make it easier to see that my smaller series is equal to infinity. They are not required for that statement about my series to be true, nor are they meant to be a mathematical part of my smaller series. The grouping is explanatory only.

[@sigmaoctantis5083]

@@babyboy55531) No rearrangement here, just an application of the _direct comparison test:_ the series written down by @TurkishKS is divergent and its terms are always less or equal than the corresponding term in the harmonic series, which hence must divergent all the more. 2) Even a rearrangement would have done no harm: series with nonnegative terms can be rearranged at will without affecting their convergence behaviour, because they converge if and only if they converge absolutely. (That theorem of Riemann you are thinking of deals with convergent, but not absolutely convergent series; those must have infinitely many positive and negative terms.)

[@HenrikhWolf]

No comment about the perfect loop? whoa

[@bonkser]

i was looking for one

[@jose000]

Yeaa

[@alifram117]

I was about to watch it for the 2nd time lol

[@joetrident]

It's been long enough for people not to care about this anymore

[@yektaagra741]

@@joetrident its the only one which caught me off in a while now, it deserves my respect at least

[@Matthew_Klepadlo]

Ah yes, the endless truth of calculus: infinitely many things that are infinitely small giving you an exact solution to a problem…

[@thomasmiller8476]

Ahhh it's relieving to know that infinity-infinity can equal anywhere between negative infinity and positive infinity. Wait no. That stresses me out. Why calculus! WHY!!!

[@zlosliwa_menda]

The video literally shows you that no, it doesn't always give you an exact solution. No, infinity isn't an "exact solution".

[@e.Kab.]

@@zlosliwa_menda Congratulations!!! _Matthew Klepadlo_ has introduced you to *Sarcasm* yayyy! 🥳👏

[@zlosliwa_menda]

@@e.Kab. Why would he be sarcastic about this?

[@Josh729J]

It gives you a very close approximation.

[@commy1231]

No calculus needed. Group up the terms like this: the first term, then the second, then the next 2, then the next 4, then the next 8. The first is greater than 1/2. The second is greater than 1/2. The next 2 are each greater than 1/4, so their sum is greater than 1/2. The next 4 are each greater than 1/8, so their sum is greater than 1/2. The next 8 are each greater than 1/16, so their sum is greater than 1/2, and so on... You get the sum of infinitely many times 1/2 which is infinity

[@MistahPhone]

Another good explanation for this problem is this: Take this equation, for example X = 1/9 + 1/10 + 1/11 ..... 1/16 All of these are greater than or equal to 1/16, so let's just say x = 1/2 for demonstration purposes Repeat this with 1/17...1/32, 1/33...1/64, and so on to get infinity because ½ × infinity is still infinity Sorry if that didn't make sense

[@m.a8335]

+ You just can't add infinite numbers - Hold my integral

[@multiversalrevan]

Series too

[@timm1328]

“put the excess area in a cap and call it Mascheroni”

[@Qermaq]

You're yanking my doodle!

[@gytoser801]

Can you prove is it rational or irrational

[@timm1328]

I have a truly marvelous demonstration that it is irrational that this KZbin comment is too narrow to contain.

[@oni8337]

@@gytoser801 You really think that can be answered right here right now huh

[@channalbert]

*”Euler-Mascheroni”, but then again, if we apply this all over mathematics, every constant and theorem would be “Euler-Someone Else”

[@starplayzreadbio373]

That one math teacher who doesn't leave you until you get the answer:

[@ntlake]

Yes, and if you rotate that graph around the x-axis the volume will be π. π ∫∞₁(1 / x²)dx = π(−(1/∞)+ ‌1⁄1‌) = π(0 + 1) = π

[@dannon9699]

“Just under two” me every time someone asks if I know my limit.

[@1996Pinocchio]

Then, the limit is 2, not "just under two".

[@simonecoppola8149]

The result is so harmonic

[@gabrielsmith3993]

Nice one 👍

[@PepeChess-vh5ef]

It's called harmonic series and with n approaching infinity the sum will also be infinite. You can find a nice proof of it on wikipedia

[@avivlab]

There is proof that doesn't require integrals, which makes it more elegant IMO. You can lower the sum by replacing some of the terms with smaller terms to write it as: 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + ... which doesn't seem so clever at first because I just substituted each denominator with the next power of 2. However, when you go to calculate this new sum, which is smaller than the original sum, you get: 1 + 1/2 + 2/4 + 4/8 + ... = 1 + 1/2 + 1/2 + 1/2 + ... and it much more obviously diverges, and means the original sum diverges as well. As a bonus, you can see the divergence of the new sum is logarithmic. If you think about it, each time we need twice as many terms to increase the total sum by 1/2, which is a logarithmic behavior.

[@Laittth]

crazy how the format of this video made the loop so seamless!!

[@fansana8641]

He is idiot

[@user-zp1xn1rm8f]

integral of 1/x = ln|X|=> which means that as x tends to infinity, the limit will also be infinity

[@AkshatIITK]

Using Calculas. Using expansion of ln(1-x) = -(1/1 + x^2/2 + x^3/3 + x^4/4 ....... ) Putting x=1 Ln(0) = -(1/1 + 1/2 + 1/3 +..........) Therefore, -Ln(0) = 1/1 + 1/2 + 1/3 +..... Inf = 1/1 + 1/2 + 1/3 +.....

[@AlejandroMGG]

There is a much easier way to see why it is infinity. We can substitute each term of the sum with 1 over the next power of 2. For example, 1/3 we change it by 1/(2^2)=1/4. Therefore, we go from 1+1/2+1/3+1/4+1/5... to 1+1/2+1/4+1/4+1/8+... affirming that the original series is greater than the second one, as 1/3>1/4, 1/5>1/8... and so on. From the second series we can group terms this way: 1+1/2+2*1/4+4*1/8+8*1/16+... = 1+1/2+1/2+1/2+1/2+... = 1+2*1/2+2*1/2+...=1+1+1+... which means that, given the series is infinite, we are adding up infinite ones. Therefore, as our original series is greater than the second one, and the second one is infinite, the original is also infinite.

[@starwarsjoey228]

me who thought the answer was 2

[@fos1451]

That's 1/1 + 1/2 + 1/4 + 1/8 .... X below is divided by 2

[@hrperformance]

@@fos1451 sweet. Thanks 👍🏽

[@georgiaoni6101]

@@fos1451 does it not work with 1/1+1/2+1/3? i kept going further and further on the calc and it did seem to just get closer to 2

[@fos1451]

@@georgiaoni6101 The answer would be bigger than 2, are you sure you put it correctly? It shouldn't be "closer" to 2, it should be bigger than 2 I just tried it and you already got the 2.08 on 1/4 and 3.01 on 1/11

[@georgiaoni6101]

@@fos1451 ah no i did it wrong i was doing 1/2 1/4 1/8 oops

[@paul_w]

Its the series/integral equivalence theorem. See: Integral test for convergence on Wikipedia.

[@AhmaD_1-9]

you are gay?

[@brad1ey50]

Yet 1-1/2+1/3-1/4+…=ln(2). Fascinating

[@alessandrolavezzicrovetto6012]

You can try Oresme's method: Use the series S=1+½+⅓+¼ + ... and the use another serie that is "minor" that S. We call this new serie G... G= ½+½ + ¼+¼ + ⅛+⅛+⅛+⅛ + ¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶ + ... As you can see, this new serie can be expressed for the following form: G= (½+½) + (¼+¼) + (⅛+⅛+⅛+⅛) + (¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶+¹/¹⁶) + ... G= 1 + 1/2 + 1/2 + 1/2 + ... G= 1 + 1 + 1 + 1 + ... In conclusion, G diverges, and that means that S diverges because a "minor" serie diverges.

[@coffeebean4939]

them: “so NOW what if i ask you this question” me: i still don’t get it edit: guys ur smart but i barely passed seventh grade math so the chances of me ever understanding this are extremely low 😂

[@Matt-wh6wj]

The point is that any series of numbers repeated infinite times, if not inversely exponential and not equal to zero, no matter how small it seems to be, is equal to infinity

[@kirito6170]

@@Matt-wh6wj i can't get it for me that looks like a 1,5 if u use rectangle

[@hellfireofdooom8876]

@@Matt-wh6wj have you heard of p-series lol

[@RanDom_GuY_yt]

@@Matt-wh6wj bro if you have something like an infinite gp with r

[@Matt-wh6wj]

@@hellfireofdooom8876 the p series is a divergent series

[@Deathranger999]

My favorite explanation for why the harmonic series diverges is this: we can collect every 2^n terms and observe that their sum is at least 1/2. For example: 1 > 1/2 1/2 >= 1/2 1/3 + 1/4 > 1/4 + 1/4 = 1/2 1/5 + 1/6 + 1/7 + 1/8 > 4 * 1/8 = 1/2 So the sum of the whole series is at least 1/2 + 1/2 + 1/2 + …, which clearly diverges.

[@bananafreyr3968]

If you replace every number by the 1/2^n inferior, you get 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16+1/16 ......... which is equal to 1/2+1/2+1/2+1/2...... and so on till infinity. Sunce the first sum is greater than the second one, the first one must approach infinity aswell 😊

[@ireallylikepizzasok]

Also, if you round down 1/3 to 1/4, as well as rounding down 1/5, 1/6, and 1/7 to 1/8, and so on, then if you add those up it's 1 + 1/2 + 1/2 + 1/2... which is infinite

[@jemuzuthestrangebear]

when I first heard about this they proved it by using a series that goes 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 ... and if you compare it to the original series 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + 1/16 ... 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 ... you can see the original series is clearly larger and you can also see the smaller series is basically just repeatedly adding 1/2 and if you add the same number over and over again it will eventually escape to infinity and since the original series is larger it will also escape to infinity

[@diribigal]

Honestly, that method is so much more straightforward than "develop enough calculus to know that the integral of 1/x grows without bound, and then compare areas"

[@themathsgeek8528]

Yes, that was by Nicolas ohrem, French mathematician.. (idk if I got the spelling right)

[@lyrimetacurl0]

I was surprised that 1/2^n sum is the same answer as 1/triangle numbers sum, even though the triangle numbers one is not exponential 🤔

[@themathsgeek8528]

@@lyrimetacurl0 hmm true, that is quite interesting... yes the triangle number problem can be done by partial fraction decomposition

[@Untoldanimations]

@@diribigal This method is cool but not generalisable. The calc method is not as cool but very easy and very generalisable.

[@gabedarrett1301]

This is how the integral test should be taught in school. This is so intuitive!

[@-_-_-_-_]

It is taught like this tho

[@halfbl00d55]

@@-_-_-_-_ it i taught, but some dont listen. Or some cases the trachers just explain ot like shit Most likely the former

[@gabedarrett1301]

@@-_-_-_-_ Maybe your teacher did it that way, but you can't assume everyone else had a teacher like that

[@ha-kx9we]

@@gabedarrett1301 that is calculus, it was taught in last year of high school to me, its litterally how it is defined.

[@gabedarrett1301]

@@ha-kx9we Well good for you, but again, you cannot assume my professor taught it intuitively like this

[@jasonpark5247]

Another explanation I like: take a series, 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8… Comparing to 1 + 1/2 + 1/3, it has to be smaller or the same. Now add 1/4s, 1/8s, 1/16s and so on. You have 1 + 1/2 + 1/2… which goes to infinity. Since 1 + 1/2 + 1/3… is larger than or equal to that, it also goes to infinity.

[@stefanomaracci7961]

Without integrals you can say that 1/1+1/2+1/3+1/4+1/5+... > 1/1+(1/2+1/2)+ (1/4+1/4+1/4+1/4)+ (1/8+1/8+1/8+1/8+1/8+1/8+1/8+1/8)+... = 1+2*(1/2)+4*(1/4)+8*(1/8)+16*(1/16) = 1+1+1+1+1+1+...= 1*infinite= infinite

[@niksforeve2805]

There is an explanation without using calculus (Cauchy's integral convergence test for sum of series): Let's introduce new series with elements each not greater than the respective initial one, precisely: {1; 1/2; 1/4; 1/4; 1/8; 1/8; 1/8; 1;8 ...} Now let's look at the sum of this series: 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+... = 1 + 0.5 + 0.5 + 0.5 ... As we see we are constantly adding 0.5 to 1, as the sequence is infinite, the process lasts infinitely long. So the sum goes to infinity and beyond, hence this series diverges. As the initial series has elements each not less than this one, then it diverges too, so it's sum goes up to infinity too.

[@nayjer2576]

Thats the proof we used in our real analysis class

[@rubenvela44]

Now this is - 1⁄12

[@SayAhh]

I thought it was -12

[@rawenmakboul7679]

That has been proved wrong, it was never true, it doesn't even make sense

[@marcosenni3110]

Clickbait and that was 1/n^2

[@ravenking2458]

@@rawenmakboul7679 Where?

[@rawenmakboul7679]

@@ravenking2458 here is the link

[@spongebobseyelashes8548]

You can prove it’s infinity due to the p series tests for sums, here it is 1/n^p where p is 1. In order for the sum to converge (not be an infinity) p must be greater than 1. Here it will diverge to infinity, which has already been proven to great lengths about this one (the harmonic series) but it’s still fun to prove yourself

[@SISKCERTWaJaVlogs]

the series diverges via the integral tests as the pseries rule states, if n > 1, the series converges and if its 1 or less then it diverges 1/n limit approaches zero but proven by the integral test it diverges

[@galaxyyy3427]

do the wallis product sometime around :)

[@jordan7828]

My nigga💀

[@galaxyyy3427]

@@jordan7828 Hello Chris.

[@augustdruzgal475]

My favorite proof of this is the comparison of the limit of 1/x as x approaches infinity being greater than 1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + 1/8 × 8 + 1/16 × 16 + ..., and since each of the chunks adds up to one and you can separate them into infinitely smaller portions, 1/x must approach infinity as well. It's just so easy to grasp visually, I love it.

[@enriquegrageda]

His loop is perfect 🫡💪

[@Nick-the-fox]

also u can do without integrals 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8...> v v v v v v v v 1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16...= 1/2+1/2+1/2+1/2+1/2+1/2+1/2+1/2+...=inf/2=inf

[@petrificustotalus9307]

Mom the video isn’t ending

[@ericsanders7226]

This is what my calculus teacher told us: 1/1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8... = 1 + 1/2 + 1/2 + 1/2 ... If we reduce everything so that the denominator is a power of two (this is less than this series), it will continuously add 1/2, and because 1/2 * infinity = infinity. Because the power of two series is divergent, the harmonic series must be too.

[@MasterHigure]

This is the solution that people who haven't finished high school still have the tools to understand.

[@ayaipeeoiiu8151]

@@MasterHigure or because it’s an other way to see the problem.

[@AalapShah12297]

Was going to comment this. Much more simpler than involving calculus

[@Prasen1729]

@@MasterHigure lol are you high on yourself ? :-) ridiculing his method saying it's for high school kids

[@ericsanders7226]

@@AalapShah12297 Yeah, that is what I was thinking. And I myself am actually a Cal B student right now and we learned improper integrals and how the harmonic series is divergent within a month of each other. So the only people who would know that the integration of 1/x from 1 to infinity is divergent and not know about the harmonic series being divergent would be a very small group of people.

[@hasanakhtar8629]

can also be expressed as integral 1+x+x^2+...+x^n from 0 to 1 and n tending to infinity. solve the limit and we get infinity :)

[@CanadaGovernment785]

Very good video I can see you put the work in!

[@BlueSheep777]

From coder's perspective: for each 10^n the sum will be by around 2,3 higher 10: 2,93 100: 5,19 | ~+2,3 1k: 7,49 | ~+2,3 10k: 9,79 100k: 12,09 1M: 14,39 Each of them is ~ +2,3 more than last 10th's power sequence. Script: var number=0; for(var i=0;i

[@KrishnaPahuja8]

Thanks for the code dude can u also do python

[@BlueSheep777]

@@KrishnaPahuja8 \t means Tab number=0 i=0 while(i

[@incizor1273]

I had a teacher who explained this to me something like: ”1 > 1/2, and 1/2 is >= than 1/2 (I mean it’s equal to but that’s not the point) and then (1/3 + 1/4) > 1/2 and after that (1/5 + 1/6 + 1/7 + 1/8) > 1/2 and you can go on and on like this and add more and more numbers that are greater than one half.” Then I don’t remember the formula but he wrote it with summary notation that you get ”n halves” I think and so the series is indeed infinite if you keep adding numbers that sum to one half (or greater).

[@nieczerwony]

Well in set of real numbers you can always find one which is smaller then the previous, so nothing magical here.

[@jbkim515]

That's how I learned. I guess it's the best explanation for showing the sum is infinite 'before learning the calculus'.

[@jay9675]

1/(1-x) = 1 + x + x^2 + x^3 + x^4...... integrate both sides from 0 to 1 -ln|1-x| = x + x^2/2 + x^3/3..... = -(ln0 - ln1) = 1 + 1/2 + 1/3.... here, ln = 0 tends to -ve infinity so, 1 + 1/2 + 1/3 + 1/4.... = infinity

[@syed3344]

Answer is correct but the method is wrong X should lie btw. -1 and 1 exclusive

[@michaeldelmundo6043]

2x(1/2+4÷0)[01]=infinity so it's bigger than infinity

[@ahmadbels]

I like how the question is asked infinitely

[@charvikdixit8328]

Thank you so much! Bri and keep making videos like this. Almost nobody makes concepts such easier.......... 👍❤

[@AhmaD_1-9]

you are gay?

[@hagenfarrell]

It’s interesting because you’d think when you take the limit you get 1/infinity (which is 0) but the infinite series of 1/x is divergent; it never converges to a finite value. It’s also known as a harmonic series. Conversely though the alternating harmonic series is convergent; it converges to a finite value.

[@babyboy5553]

Sum(1/n) is a Dirichlet series with exponent=1, therefore is divergent which means its sum its +inf

[@EliteCubingAlliance]

WTH! That loop is flawless!

[@karlkastor]

I think the proof by grouping values until they are larger than 1/2 is easier to understand.

[@franchello1105]

I like the grouping proof. First group is 1/2. Next it's 1/3 and 1/4. Then it's 1/5, 1/6,1/7,1/8. Etc. If you round each groups member to the value of it's smallest and then sum them up, you get 1/2 as it's sum, which is smaller than it's actual sum. So the modified sum becomes 1 + 1/2 +1/2 +1/2 +... , and this diverges to infinity. And the real sum must be bigger, so it diverges too.

[@Kero-zc5tc]

I just knew it as 1/3 -> 1/4 1/5 ->1/8 etc so you get 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8…. The quarters become a half the eighths become a half as well so you get 1+ 1/2+1/2+1/2…. This is smaller than that so you know it’s infinite

[@0t019]

1/1 + 1/2 + 1/3 •••• > A: 1/1 + 1/2 + 1/4 +1/4 + 1/8+ ••• A= 1/1 + {1/2+ 1/4*2} + {1/8*4+1/16*8} + ••••• A= 1 + 1 + 1 + 1 + ••••• =infinity -From Korean student

[@SuperNova88]

Omg ty Btw I don't understand the relation between area under the curve and rectangle with the function. Is it valid for every sum?

[@abhrankan1461]

An extract taken from the introduction of one of Euler's most celebrated papers, "De summis serierum reciprocarum" [On the sums of series of reciprocals]: I have recently found, quite unexpectedly, an elegant expression for the entire sum of this series 1 + 1/4 + 1/9 + 1/16 + etc., which depends on the quadrature of the circle, so that if the true sum of this series is obtained, from it at once the quadrature of the circle follows. Namely, I have found that the sum of this series is a sixth part of the square of the perimeter of the circle whose diameter is 1; or by putting the sum of this series equal to s, it has the ratio sqrt(6) multiplied by s to 1 of the perimeter to the diameter. I will soon show that the sum of this series to be approximately 1.644934066842264364; and from multiplying this number by six, and then taking the square root, the number 3.141592653589793238 is indeed produced, which expresses the perimeter of a circle whose diameter is 1. Following again the same steps by which I had arrived at this sum, I have discovered that the sum of the series 1 + 1/16 + 1/81 + 1/256 + 1/625 + etc. also depends on the quadrature of the circle. Namely, the sum of this multiplied by 90 gives the biquadrate (fourth power) of the circumference of the perimeter of a circle whose diameter is 1. And by similar reasoning I have likewise been able to determine the sums of the subsequent series in which the exponents are even numbers.

[@rajdeepnaha1242]

Another way to do it will be to put -1 in the expantion of ln(1+x)

[@Eitri-TheSus]

i feel like theres an easier to understand way to prove this, like the fact that the sum is larger than 1/1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8... which can be simplified to 1 + 1/2 + 1/2 + 1/2 + 1/2... which is infinity

[@adios8919]

YOU JUST EXPLAINED THIS BETTER THAN ANYTHING I'VE EVER FOUND. OH MY GOD THANK YOU SOOOOO MUUCHHHH 🤩🤩🤩🤩

[@kingju1ius]

Im in grade 10, and i can barley do math 7 grades below my level, but this was cool.

[@l.7428]

From which country you are ?

[@sohamadak6811]

Not Indian Ig

[@jayxone]

Another way, around down everything so it’s equal to any number of 1/2^n, you will see that it would be 1 added till infinity, since this is infinite and the sum is rounded down, the actual equation would be bigger than infinity or just infinity

[@Lost-soul1383]

need more pages like yours on social medias

[@orangesite7625]

Being a calculus student it really helped me understand there is a lot can be done with calculus

[@aayushroka5243]

People: Perfect loop doesn't exist This short video:

[@kavinesh_the_legend]

Teacher: There is nothing greater than infinite Vsauce: OR IS IT!!!

[@amirraza6055]

I just thought it was Geometric Progression and I felt trippy for a minute!

[@hrperformance]

The reasoning seems obvious when it's told you well like this, but I struggled a lot with this one haha. QUESTION: why was the variable t introduced in the final expression? Shouldn't this just be x, with x tending to infinity?

[@justcommenting5117]

Since the function 1/x already uses x, t was used instead to occupy the limit

[@parijatgoyaliitkgp7572]

Another way that is coming to my mind: if we write it as ∫(1 + x + x^2 + x^3...)dx from x=0 to x=1 , we get the same series, ie 1+1/2+1/3... solving, =∫ (1)/(1-x) from x=0 to x=1 (using infinite GP formula, and here common difference

[@juliusczarnecki3992]

What's funny though, is that if you revolve that area around the x axis and take the volume, you get a finite area (that is a multiple of pi if i remember correctly). Just another reason infinity isn't a number

[@iamsheel]

You start with 1,after 3 sums you got a 2.08. So after many many sums you keep getting bigger numbers until you reach infinity.

[@williamgomez6226]

I end up confused, just like in class

[@karlmudsam2834]

This was literally like “do the integral test lmao”

[@bill_exists]

the loop was so clean he be asking the same question infinite times.. 💀

[@madhavan8195]

Int 1/x dx under limit 1 to infinity, log|x| under limit 1 to infinity, log|♾️| -log|1| , # log|♾️|= ♾️, log 1 = 0 log|♾️| - 0 ♾️ - 0 = ♾️ Hence, int 1/x dx under limit 1 to infinity is ♾️ .

[@mehulmishra7630]

This can be understood from the Machlaurian series expansion of Log(1-x) .. putting x = 1 which gives the answer infinity.

[@lifestreamy]

Help stepmathbrother, I'm stuck in a loop

[@purplegoldcat7]

Smoothest transition does exist now

[@gigantopithecus8254]

you could prove ln(x)-h(x)=v as x approaces infinity which makes it trivial

[@1SLMusic]

Here’s best proof for this which doesn’t involve calculus. Begin with the sum: 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + … We can compare it to a different sum: 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 + …. Because of how each term increases to the next, we know the second infinite sum is less than the first infinite sum if both have integer solutions. This means that if we can prove that the second sum evaluates to infinity, then we also prove that the first sum also evaluates to infinity. We can simplify the second sum as follows: Sum = 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + …. Sum = 1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + …. Sum = 1 + 1/2 + 1/2 + 1/2 + … Thus, the second sum is an infinite sum of 1/2. which is infinity. Because the second sum is equal to infinity, the first sum 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + … = infinity.

[@tomekk.1889]

Calculus is way faster

[@bruhyou197]

This is amazing.

[@josemarino8787]

The sequence (s_n):s_n=1+...+1/n is not a Cuachy sequence since s_(2 n) - s_n>=1/2. Hence, (s_n) doesn't converge. Since is increasing, s_n must go to infinity.

[@jaynk4395]

1))this is kinda fun if we make a function of this as f(X)=1/x and put limit x tending to infinity ... then by d sub we get 1 over infinity which is tending to 0 or infinitely small 2)) if we integrate the function for all real values we get a function log(x) and if we put limits in it it will be log of infinity = infinity 3)) if we try to understand it with the help of logic that the values in denominator are increasing serially and a natural no .... then by seq n series ..... sum of all natural no = 1 over n(n+1)/2 and put n as infinity since there are infinite natural no and apply summation then we got infinity over infinity which is infinity just so many ways

[@kingbeauregard]

It clearly sums to -12.

[@ravenking2458]

No dude, it's 1+2+3... = -1/12(Atleast that's what Ramanujan stated)

[@AdityaGupta-qk5ee]

@@ravenking2458 ramanujan stated that for 1 + 2 + 3 …….. this is 1/1 +1/2 +1/3 ………..

[@ravenking2458]

@@AdityaGupta-qk5ee yep, sorry i forgot to mention that. I meant that he probably did reciprocal of the whole series of 1+1/2+1/3....

[@RaRa-eu9mw]

The reciprocal of a series isn't the series formed from the reciprocals of the terms. 1+2+3+4...=-1/12 (if you want to give it any value) but 1+1/2+1/3+1/4...=infinity (if you want to give it any value)

[@ravenking2458]

@@RaRa-eu9mw Dude i said the same thing, he probably did the whole reciprocal of the series which was obviously wrong

[@zjz1]

You don't need calculus. 1/3+1/4 > 1/4 + 1/4 = 1/2 1/5+1/6+1/7+1/8 > 1/8+1/8+1/8+1/8 = 1/2 and so on The sum > 1+1/2+1/2+1/2... = Infinity

[@nicezombie8054]

There are way easier proofs, for example you can easily figure out that you can divide it in an infinite amount of areas with a size bigger than one half

[@soumyadeepdash4805]

Yes infinity. If this taken upto n terms, the sum will be approximately log(n)

[@miamitten1123]

The loop is infinite ♾

[@AhmaD_1-9]

you are gay?

[@HamboboCute]

1+0.5+0.3(point three repeating)+0.25>2,so at least 2

[@dercivity3870]

now subtract the area from the rectangles and you get 0.577. Math is awesome

[@johnshillingsburg2682]

This is the harmonic series. Currently in AP calculus BC and learning sequences and series. This is the number 1 fact to know.

[@maicee7603]

how is the area of the rectangles represent the sum? is it because its the same as sum of 1/x ×1 from x to inf?

[@emilyscloset2648]

The first rectangle is defined with an area of 1, the second is half its height so 1/2, the third a third of the first rectangle's height so 1/3 etc

[@methatis3013]

Yes

[@hydrochloricacid2146]

Look at it this way: the area of each rectangle is simply the rectangle height (1/n) times the base. Because we're using indices going 1,2,3,4 etc we're actually looking at a base of width 1. Adding up all those fraction is this equivalent to adding up those areas.

[@zackcarl7861]

Its infinite , that's like , that elephants father can fit in the shade, and that, elephant is big how big it's the biggest , but then what about his father he is bigger than that biggest elephant , so do we have have enough shade space to fit the elephant 🙂😆

[@redeaglee]

Round all fractions to powers of 2 so 1/3 is 1/4 it would work because it's kinda of like taking 1/5 into 1/4 we have an extra so we put it to 1/3, then add allnof the equals and tou will get 1/2+1/2+1/2_ so it is infinite