I heard 1/2+1/3+1/4+... goes to infinity but I didn't think it would go past 3. Reddit r/learnmath

[@bprpcalculusbasics]

"If you add infinitely many positive number you get positive infinity"? In fact, it’s not! kzbin.info/www/bejne/eYDIinavZsmeg9E

[@Bla_bla_blablatron]

is this what we suffered for? To be held down by systematic racism! We Democracy. We vote now. Gibs for all Under privileged inner city people!

[@Bla_bla_blablatron]

...and sheeeeet

[@netdragon256]

I didn't watch this, but it's easy with calculus. The integral of 1/x is ln |x| + c which goes to infinity, as x goes to infinity, albeit very slowly. Sure, your summation formula is integers, and the integral is more granular, but it doesn't matter because the CARDINALITY IS THE SAME as you approach infinity! One just grows faster than the other. Obviously, this isn't as pleasing because it depends on the understanding of the cardinalities of the sets are identical, however we know this to be the case.

[@Zorgdub]

That's true if you're adding integers :P

[@spiffyavatar3611]

😊​@@Bla_bla_blablatron

[@vladimirpotrosky7855]

Marker switches so smooth

[@kaustubhgupta168]

yea hes very good at it

[@bprpcalculusbasics]

Thanks!

[@EaglePicking]

I ignored the math completely and was just mesmerized by his sleight of hand color changes.

[@diggoran]

And now you know why the channel is named bprp, because it stands for black pen red pen. The switching is part of the branding.

[@EaglePicking]

@@diggoran ***mind blown***

[@VeteranVandal]

It's just slow, but it goes "to infinity" through sheer fraction will.

[@hardboiledegg2681]

the sheer fraction will when I introduce 1/n^2

[@obiwancannoli1920]

The indomitable fraction spirit

[@Saber1320]

@@hardboiledegg2681 well doesnt this just converge to pi^2/6 ?

[@hach1koko]

@@Saber1320 yeah that's the point, this one would converge despite "sheer fraction will"

[@jayi31]

@@hach1kokothe futile fractional will vs the unstoppable power of a square

[@crclayton]

I haven't done math proofs or calc. since college but KZbin has correctly started promoting this content to me and it's really nice to see these clean demonstrations.

[@KCJbomberFTW]

Same I’m an accountant I got into school with this 😅

[@Bla_bla_blablatron]

math is Racist! Y'all betta go back to da Crackerbarrel

[@pedrogarcia8706]

probably means you share interests with college calc students!

[@saravanarajeswaran2626]

Can you teach a proof why pi is irrational, like assuming it to be rational and prove it with contradiction

[@Qermaq]

Mathologer has done a really good video on that. It's quite a bit more challenging than this problem.

[@maxgoldman8903]

kzbin.info/www/bejne/mneuk4mKfbBnqrcsi=n08QZRrYRXzSRdfM

[@customlol7890]

That exercise is trivial and left as an exercise for the reader - Fermat probably

[@maxgoldman8903]

kzbin.info/www/bejne/mneuk4mKfbBnqrcsi=01bxPVz8EVOpP7E-

[@saravanarajeswaran2626]

@@Qermaq Yep i saw in mindyourdecision too, but they're seen to be complicated (even though he showed that easiest proof), so i asked him to do, because it could be easy to see

[@trillionbones89]

Answer: It approaches infinity incredibly slowly.

[@EhabShahid]

*infinitely slowly

[@bohanxu6125]

Logarithm: "hold my beer" ... "It's going to take awhile"

[@chickenlittle8158]

... can things approach infinity any slower or faster than something else?

[@derblaue]

​@@chickenlittle8158Actually yes. If we take the partial sum and compare them for different series we can see the different "divergence speed" by the ratio of the partial sums as we increase the terms of the partial sums.

[@burgundyhome7492]

Infinite numbers, just like finite ones, are not equal and comes with infinitely different sizes.

[@worldnotworld]

This proof is due to Nicole Oresme, a medieval mathematician, theologian, and philosopher.

[@nchappy16]

mfs really just had “smart” as their job and worked in every educated field

[@worldnotworld]

@@nchappy16 So true. And the fields weren't as distinct from each other then.

[@A38]

​@@nchappy16People have that job today. There are even people whose job is just to communicate breakthroughs in science and math to the layman. The difference between the most and least educated humans is EXTREME.

[@bardsamok9221]

Would be good if he proved God

[@dovallebr]

@@bardsamok9221Saint Thomas Aquinas already did that tho

[@o_s-24]

I hate this fact. This series has business being divergent, but here we are

[@9adam4]

Clearly the series should be exiled from the system and mount a resistence. 😊

[@pixynowwithevenmorebelkanb6965]

A bunch of russian mathematicians go to a bar.. The first one wants one bottle of vodka, the second one wants half, the third one a fourth and so the others. Eventually the bartender puts 2 bottles of vodka and exclaims: Sort it amongst yourselves!

[@Misteribel]

The beauty of this series is that it doesn't work in 3d: the paradox of Gabriel's horn. If you turn the graph around is x-axis to get a nice horn, the total content of the horn is NOT infinity, but converges to π.

[@vanshamb]

π and e are omnipresent man

[@galoomba5559]

​@@vanshamb There's circles involved. Not that weird that pi appears.

[@vanshamb]

@@galoomba5559 ohkk then just e is omnipresent ig

[@CliffSedge-nu5fv]

Infinite surface area, but finite volume. You could fill it with paint, but can't paint its sides.

[@alinasar6192]

​@@CliffSedge-nu5fv daym that makes no sense, maths wrong

[@matta5749]

Fun fact: even if you only have inverses of primes (I.e., 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + … ) it still diverges to infinity. But if you take out all denominators that contain a 9 from the normal harmonic series (I.e., 1/2 + 1/3 + 1/4 + … + 1/8 + 1/10 + … + 1/18 + 1/20 + … ) it converges to a finite sum. (Credit to that one mathologer video)

[@matc241]

Lol, that second one seems counterintuitive but actually makes sense thinking about it.

[@sdfsdfhgcvbn2425]

​@@matc241why would the second make sense?

[@RCmies]

I don't understand

[@theeraphatsunthornwit6266]

How come the second one converge. . DOES NOT make sense at all. I dont believe it. If that one converge, then the took out part must also be converge coz it is smaller , then recombine them , must result in converge value.

[@timonbubnic322]

@@theeraphatsunthornwit6266 i agree yeah, it also doesnt make sense considering there are more multiples of 9 than there are primes...

[@user-nd7rg5er5g]

I feel like I'm a toddler being successfully convinced to enjoy eating vegetables by a knowledgeable older person. Math normally slides off my brain like water off the back of a duck, but this explanation style was excellent, and taught me a lot!

[@TheAlienPoison]

This comment is misleading, I still don't enjoy vegetables after watching the video.

[@KAMUPhobies]

she toddlers on my vegetable til i math

[@thecrazygamertarun5265]

Goo goo gaa gaa daada baabaa

[@Matt-lv1jl]

Bruh how can one not enjoy vegetables, that stuff is addictive af@@TheAlienPoison

[@fernando4959]

​@@Matt-lv1jl dog what veggie you consuming

[@Joefrenomics]

Ah, darn it! I tried to prove it myself first. Failed. Then saw your proof and thought “Darn it! Why didn’t I think of that?!”

[@bprpcalculusbasics]

I didn't come up with these proofs myself. I learned them when I was a student. I most likely wouldn't be able to come up with them myself.

[@MikehMike01]

chocolate pudding

[@rhel373]

@@bprpcalculusbasics My maths knowledge doesn't go beyond... I guess high school level, and I don't really have any clue how a proper mathematical proof works. To me, before having finished the video, it seems like any sequence where the number added doesn't eventually become zero should approach infinity. There's always *something* added so the number will keep growing for infinitely many steps. Given how simple that sounds I can only imagine I'm making a mistake in my logic there, because usually these things are more complex my intuition tells me.

[@tjaschicksalVegita]

​@@rhel373 If you take the sum of 1/(2^n) = 1/2 + 1/4 + 1/8 + ... It approaches 1 not infinity, even though a number > 0 is added to it infinitely times

[@psymar]

Understanding a proof is generally much easier than coming up with said proof

[@GynxShinx]

Very funny to think that the whole front of the harmonic is finite no matter where you cut it off so if you start at 1/1000000000 and then continue the harmonic series +1/1000000001+1/1000000002... and so on, it still diverges to infinity.

[@CesarDainezi]

I saw this demonstration in my Real Analysis class 10 years ago. Very elegant.

[@cbunix23]

I saw that proof in middle school.

[@CesarDainezi]

@@cbunix23 In middle school I didn't now what x meant LOL

[@zigzagnemesist5074]

@@cbunix23You must live in some advanced alien civilisation if you’re doing these proofs in middle school…

[@cbunix23]

@@zigzagnemesist5074 Haha. It was a long time ago; it might have been first year high school. Heck, my dad did spherical trigonometry in high school 100 years ago.

[@AnnXYZ666]

​@@cbunix23Sheet, what kind of alien civilization taught spherical trigs in high school even before ww2?

[@killer100897]

I didn't noticed his second marker at first and got confused on how the color suddenly changed without putting down his arm 😂

[@digitig]

You didn't know what the channel name, bprp, stood for, did you?

[@killer100897]

@@digitig nope just saw this video in my recommendation and decide to click it on a whim

[@eduardoteixeira869]

Thank you. I knew this is a divergent series, but I did not remember if I saw a demonstration for that. Thank you to show me such elegant proof.

[@Rurumeto]

It approaches infinity very painfully

[@philippenachtergal6077]

An intuitive (not rigorous) way to see this goes like this: If you take all elements between index 10 and 100, there are 90 of them and they have a median value of 1/55 so you expect the sum of them should be above 1 If you take all elements between index 100 and 1000, there are 900 of them with a median value of 1/550 so you expect the sum of them to be above 1 And you can go on like that for infinity. The above is not a proof because the median value can be bigger than the average value but it is intuitive and you get the feeling that there has to be a proof that works "kind of like that". And indeed, its not hard to change that into a proof as each of the steps above can easily be proven to have a value > 0.9 by replacing all its terms by the last term. This is quite similar to the usual proof made in the video.

[@WhiteGandalfs]

Essentially, that's the same as he does but with base 10 instead of base 2. I think this one is better suited for children that are not yet familiar with base 2 thinking.

[@AbramSF]

He didn’t take the median value in this proof with 1/2 though. Each power of 1/2, he took the minimum value. Which allows any sum of those numbers to be bigger than the minimum.

[@renem.5852]

Just don't look at median value, but the minimum value of your intervals. If you have 90 * (sth that is at least 1/100), you have at least 90/100 = 0.9. If you have 900 * (sth that is at least 1/1000), you have at least 900/1000 = 0.9. ... Even though I don't see how this is more intuitive than taking the powers of 2, taking an obvious lower limit should be more obvious than the median for most people.

[@philippenachtergal6077]

@@renem.5852 Of course, and that's what I explain at the end. I was showing a thought process, first I think about adding terms in an exponentially increasing batch size and I use the median value to get an idea of the partial sum because that's what first came to mind and it seems "about" right. This is enough to tell me that there is probably a proof not for away. And you get an actual proof by using the minimal value rather that the median(that doesn't bend well to prove a boundary value) or the average (because you can't easily figure out the average value). When you explore a problem, you don't necessarily get to a proof right away but misses can still be good enough to confirm some ideas and guide you toward a proof.

[@williamcomer6788]

Great video demonstrating the proof but we should make sure to make the distinction that infinity is not a number. Infinity minus a real number is meaningless.

[@fafflerproductions]

I'm gonna be honest, I just watch for how quick you swap marker colors

[@JayMaverick]

Brute force infinity.

[@matteoblasoni8726]

I have a much more simple demonstration (beware: joke ahead) -> if you have 1/∞ and sum it up an infinite amount of times, you roughly have ∞/∞ so around 1. If you keep up with the sum, you can basically sum 1 an infinite amount of times, which..... Bring the sum to infinity!! 😂

[@bitonic589]

Math is crazy. 1/infinity ≠ 0

[@magicmulder]

The sum over 1/n^s goes to infinity for s1.

[@SteveThePster]

Nice vid. Maybe worth mentioning that the numerical proof is much more powerful as it doesn't rely on the huge amount of calculus behind the theorum/proof for Integral (1/x) = log x

[@Bla_bla_blablatron]

Y'all stole that from Africa! Or maybe you never heard of Wakanda.

[@snorman1911]

​@@Bla_bla_blablatron we wuz kangs

[@BenCritchlow]

There is quite a neat proof by contradiction I read in a journal: Assume the harmonic series converges to some real number S. Then S = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ... > 1/2 + 1/2 + 1/4 + 1/4 + 1/6 + 1/6 + 1/8 + 1/8 + ... (Here all we did was replace 1 with 1/2, 1/3 with 1/4, and so on). Then, summing the halves, the quarters, sixths, and so on: = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = S We have our contradiction S > S, since no such real S exists.

[@tomkelley4119]

The first proof that you showed is really beautiful! Thank you!

[@mechbfp3219]

Even getting to a million would take an incredibly large number.

[@Hypernova7777]

Because the amount for each next ½ doubles it would take 2^1999999 for the *final ½ alone*

[@Ninja20704]

In fact, the harmonic series grows very similar to the natural log. For sufficiently large n, 1+1/2+1/3+…+1/n ≈ ln(n)+gamma where gamma=0.5772… is the euler-mascheroni constant. So the amount of terms needed to pass a given number N is approximately e^N

[@matheusjahnke8643]

@@Ninja20704 furthermore... We can approach ln(k), whenever k>1 as 1/n + 1/(n+1) + .... + 1/[kn] as n -> infinity; Round kn in the way you feel best if it isn't an integer;

[@ishansh0077]

'e' is omnipresent

[@GarrettRoyce]

@@Ninja20704 I ran it through WolframAlpha just to see what kind of numbers we're looking at here... e^1,000,000 ≈ 3 × 10^434,294 and the series from 2 to 3 × 10^434,294 is ≈ 999,999.5

[@aneeshbro]

Keep up the good content man. You are very regular, and I like that!

[@customlol7890]

Ayo the pfp?

[@aneeshbro]

@@customlol7890 coincidence

[@Bla_bla_blablatron]

you #1 Racist

[@thezanycat]

Haven’t seen the second integral proof before - really like it! Great explanation

[@superuser8636]

Shout out the limit comparison test and integral test (in order according to powering). This video is literally amazing

[@spencergrover6886]

Learned this in calc class yesterday. I think my phone is listening into my classes.

[@goldencinder7650]

if you start the video at 5:49 and playa it in reverse with the audio set to anything but your native language and that's how I was taught math in public school

[@colinjava8447]

Well S = 1/1+1/2+1/3+...+1/n > ln n, and ln n diverges, so so must S. It does require proof that S>ln n though.

[@wes9627]

Nice. Making something littler BIG.

[@tiagoandradedeoliveira8703]

man, this is just awesome. one of the best calc video ive ever seen, please, never stop doing this

[@MrRabix007]

When i saw the miniature i thought that video is about Divergence of the sum of the reciprocals of the primes . 1/2+1/3+1/5+1/7...+1/31 but it is complicated to prouf

[@Bedogg25]

I hated math but watching you explain these things is actually nice to watch

[@THICCTHICCTHICC]

This is an awesome way to explain it. I already understood it but you made it so clear. Thank you

[@bprpcalculusbasics]

Thank you!

[@Bla_bla_blablatron]

y'all stole that from black folks

[@fluffyfang4213]

Upon seeing the thumbnail (but before watching the video), I ended up doing similar logic but in multiples of 10 instead of 2. So like, set all the fractions from 1/2 to 1/10 to equal 1/10 each. Add them up and you get 9/10. Do the same for fractions 1/11 to 1/100, setting them all to 1/100 and you get 90/100... or 9/10. Glad to see I was on the right track even if going by multiples of 2 is way easier to explain!

[@indigoziona]

You somehow made it sound simple *and* made me feel smarter just watching 😊

[@nechitamarius1]

As someone who loved math in highschool and just got by in university, this is beautiful and made me feel what i felt in highschool

[@vikramvilla]

Awesome. For a series to give a finite sum, it must converge. HP doesn't converge, so the sum is infinity. Your proofs are really cool.

[@tyuh860]

What is converging? I only started line functions in math right now, so I haven't had this yet.

[@vikramvilla]

@@tyuh860 converging means if you keep on adding the numbers in a series, it shall get closer and closer to a finite number.

[@Regian]

Converging is like mathematical edging

[@jameshart2622]

Yep. If a series converges, then if you specify a tolerance, you can find a finite index where every number after that index is within the given tolerance of the true limit.

[@conorslater8295]

What a beautiful proof, thanks for sharing.

[@operator8014]

I remember seeing a neat proof my calc 2 prof showed us, but I don't remember what it was. I think he calculated the rate of shrinkage of the terms, and compared it to the rate of expansion of the sum, and showed that it remained positive for any given number of terms, therefore it must go to infinity.

[@SG-lighthouse]

I’ve seen this series explained so many times but this is the clearest. Thanks!

[@MikeGz92]

I don't remember this demonstration, but it's great 🤩

[@KaikyAlmeida-b8t]

​@@forbidden-cyrillic-handle lolll it must be italian

[@clickoppotamus]

I was having a normal day, now panicking about “how can anything ever converge???”

[@aldodzb]

If you and me are 2 meters away. And I take steps equal to half the distance between you and me, then I would never get to you.

[@Mike__B]

I just noticed on your shelf that you had boxes and boxes of markers, do you get a good price on those? As a teacher who has had to buy his own markers that looks like an absolute gold mine!

[@bprpcalculusbasics]

I got them on Amazon, ranging from $10 to $13 a box. Not a bad deal in my opinion. Btw, the “bullet tip” is the must! www.amazon.com/dp/B00006IFIN/ref=cm_sw_r_as_gl_api_gl_i_dl_TPD5KRK4A8PDMT9Y1PRR?linkCode=ml2&tag=blackpenredpe-20

[@Mike__B]

@@bprpcalculusbasics Thanks for the advice, I'll give the bulllet tip a try, I've exclusively used the chisel tip and yeah you gotta hold it at the right angle to write nicely.

[@bprpcalculusbasics]

Right. And for the bullet tips you won’t need to worry about that!

[@lbadplay]

A series is a sequence on its own, so first statement is a bit questionable

[@mtaur4113]

Idea number one generalizes to Cauchy's Test for monotonic series. The series of a monotonic sequence a_n converges if and only of the series of 2^n a_(2^n) converges. The harmonic series is a model case for the test. As a bonus, you could show that the integral of ln diverges, even if you did not know that it was the inverse function of exp, using you method 1, or Cauchy's Test on the the associated series.

[@Qcattee]

A very good and clear to follow video but also can we give credit to how flawlessly and quickly the switch between black and red marker is! I’m very impressed!

[@clintonweir7609]

I was thinking about it. For the sum of the series 1/n for n=1...N, you can create a fraction with a denominator that is N! and the terms are, I think, N!, N!/2!, N!/3! etc. And then I realized that I was just working my way back to the original problem. Whoops!

[@jameshart2622]

That is very common in math. The number of attempted derivations I've had where I ended back up at x=x.

[@oyuyuy]

I think a valuable and quick thought experiment is thinking about the sum between '1/1000... --> 1/3000' for example. It' adds up to ~1.00 which is not an arbitrary amount. Same for 1/10.000... --> 1/30.000

[@Maths_3.1415]

Harmonic Series :)

[@MyEyesAhh]

This is basically a proof via a direct comparison test. If b_n > a_n and b_n diverges, a_n must diverge. Never thought of it so much, i just accepted it via the p-series test, where p=1 (the exponent on 1/n) means it is divergent. Maybe you could do a proof where p

[@MK-13337]

Your comparison test is the wrong way around. If b_n >= a_n and a_n diverges then b_n diverges.

[@hengry2]

1+2+3 converges at -1/12 but 1/2+1/3+1/4 diverges, something is clearly wrong here.

[@ConManAU]

The sum of the integers doesn’t converge under normal circumstances. Under a different definition of convergence or of summation then sure, but there’s a good chance that the harmonic series converges in that system too.

[@justinariasluna8731]

@@ConManAU One definition/system used for these types of convergence is the analytic continuation of Riemann's zeta function (zeta(s)) to the complex plane. In this definition, we have indeed zeta(-1)=-1/12. Funnily though, 1 is a pole of zeta(s), and zeta(1) happens to be the harmonic series. So also under this definition the harmonic series is not defined!

[@danielyuan9862]

Saying it's equal to -1/12 is one thing, but saying it converges is the most illegal thing I've heard.

[@theeraphatsunthornwit6266]

A guy on mathologer channel show -1/12 thing is wrong, basically

[@ДмитроПрищепа-д3я]

@@theeraphatsunthornwit6266 it's not wrong. What's wrong is saying that the sum converges to that value. It can ve regularized to that value tho.

[@renesperb]

A fast way to do it is to note that 1/2+ +1/3 + ....+1/n > integral of1/x from 2 t0 n = ln n -ln (2) -> inf. for n -> inf.

[@Itsallover57]

Man, this is why I watch these even though Im not a math lover. That last bit was the first time someone showed me the point of an integral.

[@pangeo8183]

We can also try to calculate it straightforward. Sum 1/n from 2 to inf = Sum 1/n from 1 to inf - 1. 1/n we can replace with integral of x^(n-1) from 0 to 1. Now we have Sum 1/n from 1 to inf - 1 =sum from 1 to inf of integral x^(n-1) from 0 to 1. - 1. And then we switch the order of summation and integration, so we get integral from 0 to 1 of sum x^(n-1) from 1 to inf - 1 = (after reindexing) integral from 0 to 1 of sum x^n from 0 to inf - 1, which is a geometric series. So we get Integral of 1/(1-x) from 0 to inf - 1 = -ln(1-x) evaluated at one and zero. Then we take a limit as x approaches 1 and get -inf and 0 when x is equal to 0. So finally we have -(-inf) - 0 - 1, which is just infinity.

[@KolMan2000]

Simple answer I’d try to give if the person could comprehend is that even if you are adding an infinitely small number to another number… you are still adding a number

[@raimundo116]

I just smiled after I got it, when he added the 1/4s to 1/2s. Great explanation! Makes me remember the fun in math

[@adw1z]

Fun fact: H(n) := 1 + 1/2 + 1/3 + .. + 1/n. Then, lim n->∞ [H(n) - ln(n)] = γ , where γ = 0.577216…. is the Euler-Mascheroni constant. This shows that the harmonic series H(n) grows asymptotically like a shifted up log function, which is indeed (albeit very slowly!) divergent as n->∞. We can use this to estimate H(n) for large n, via: H(n) ≈ ln(n) + γ + 1/(2n) + O(1/n²) (this is the asymptotic expansion of H)

[@methatis3013]

Also, an interesting fact, it isn't known whether γ is even rational or irrational

[@zachariastsampasidis8880]

Wasn't it shown than γ and e^γ are algebraically independent

[@EastndFTball]

Sweet video dude, thanks for sharing. That was cool and informative

[@TheMasterGreen]

another reasoning for the geometric proof is that since 1/x has a horizontal asymptote at 0 it never reaches 0 but it approaches it as x becomes larger and larger. Therfore the "infite" term would be slightly greater than 0 (also known as nth term test) so the final series or sum will not converge because you are always adding to it forever.

[@ironeche3702]

What about sum( 1/2^x) x→inf

[@ItsLtime]

I normally would prove this via the integral test, but this one is so much simpler and honestly better!

[@danteeightsix]

I understood 1/2 + 1/3 + 1/4 + 1/5... of this.

[@stephenbachmann1171]

So you understood more than a 100% of this?

[@its-me-oni]

​@@stephenbachmann1171bro he means just the starting😂

[@alessiodibella2003]

THIS FINALLY MAKES SENSE TO ME! With the demonstration with integral calculus, I could not be convinced that 1/2 + 1/3 + 1/4 + ... could diverge, thank you very much!

[@BPGHchess]

Neat

[@delirious8424]

Hey,im 1 term stud, and this really helped with understanding of the fact that integral is S under the graphic and some other things,thanks.

[@bprpcalculusbasics]

Get your indeterminate cat t-shirt: 👉 amzn.to/3qBeuw6

[@kingjulian1202]

Why am I watching maths at midnight 😭😭

[@partyfists]

I find the latter proof slightly dissatisfying because all the proof hangs on the fact that ln(infinity) = infinity. I wish there was time taken to derive that value, or show that it is obviously infinite.

[@vijay_veluguri]

ln(infinity) is basically asking: "What power must e be raised to in order to equal infinity?". Obviously, the answer to that is infinity. Similarly, ln(2) is asking: "What power must e be raised to in order to equal 2?". And as he said in the video, that value is about 0.693.

[@partyfists]

@@vijay_veluguri yes precisely, I would have loved for that to be in the video!

[@lawrencejelsma8118]

​@@vijay_veluguri... Actually it was ln(infinity) power for e. e^(ln(infinity)) is infinity as is e^(lnx) = x substitution math that is used to especially solve Lambert W problems.

[@ishansh0077]

​@@partyfistsI mean in general the audience who watch this channel knows the reasoning behind this.

[@barnabusowl4252]

Even more peculiar, if you revolve the graph around the x-axis and then evaluate the area formed by the curve from 1 to infinity given by the integral (pi/x^2) dx, you get the finite area of pi 🤔

[@leemason4024]

Your explanation or addition of the 1/2s to get ~3 at around 5:15 in the video is mistaken. You misaligned the "1/2" for the next power of 2 with the 1/31 group which means you'd have to go to the 1/63 group to actually get ~3. As written on the board, you only had 2.5... unless I'm totally not getting this

[@Lonely_Wiz]

did you skip the video? he showed the sum from 1/2 to 1/31 at 1:19 at WolframAlpha. It is more than 3. And his explanation is on point, because he said that the sum from 1/2 to 1/31 was bigger than the sum of the powers of two.

[@coldandafraid]

That geometry proof is nice

[@vrchhu4066]

Another way to think of would be 50% + 33~% + 25% + 20% + 17~% + 14~%... I just think it's easier to visualize the infinity like this, since you see the sum increasing numerically

[@robertmayster7863]

You, Sir, have my upvote

[@mccask]

Taps the board, it magically erased!

[@TheHunter-gw7rh]

I just conceptualize it like this. 1 over any positive integer > 0. Therfore, every new fraction added to the series will increase the sum. Therefore, adding infinite fractions must increase the sum infinitely.

[@yayer_27]

this is a genius and very well-explained proof good job once more

[@bprpcalculusbasics]

Thank you!

[@BurntPencilShavings]

Uhm Just use he harmonic progression sum Where a = 2, d = 1, n = infinity We'll get the sum of the sequence as ln(infinity) = infinity Harmonic sum: (1/d)ln[(2a + (2n-1)d)/(2a-d)] Easy :)

[@tflcn]

Infinite amount of mathematicians walks into a bar. First orders 1 pint of a beer, second orders 1/2, third 1/4 etc... barmen silently pours 2 pints

[@kpk1171]

Why was I never taught this way of proving the harmonic series diverges! As well as an intuitive understanding for the integral test!

[@kckcmctcrc]

This is a brilliantly simple proof.

[@Benny-the-cute-kitty]

As a current 8th grader, I am SUPER HAPPY that you included the integral example! I couldn’t quite understand the first explanation but the second one made me understand in a split second! I LOVE that you include more than one way to solve equations! Thank you!

[@nguyenbao8987]

Not the 8th grader understanding integration... Aus education is doomed

[@Benny-the-cute-kitty]

My Turkish father taught me some simple precalculus 3 years ago, I can still remember lots.

[@Iggy_Dogg]

man I wish you were my TA when I was taking the calc courses

[@occamraiser]

interesting. But the most amazing thing is the presenter's mental agility in swapping pens seamlessly and flawlessly!

[@flameofthephoenix8395]

The most simple explanation I've thought of is to simplify it into a series of repeatedly adding 1/2 to itself, let's start with the fraction 1/2, this alone has increased the sum by 1/2, but now let's look at the fraction 1/4, to get from 1/2 to 1/4 we also need 1/3, 1/3 is greater than 1/4 so we'll simplify it to also being 1/4 meaning that there are actually two 1/4s, meaning we have to add it twice which accumulates to 1/2 again, this leaves the sum at exactly 1 in this simplified scenario, so let's jump over to 1/8 now, this time there are three other fractions that we have to pass to get to 1/8, they are 1/5, 1/6, and 1/7 we'll simplify these all to 1/8 which works since they are all greater than 1/8 meaning we're not increasing what the sum should be but rather decreasing it, however this means that there are four eights which adds up to 1/2 again, this means that by 1/8 the actual series will be greater than 1.5, this just keeps going on, every time the number of fractions added doubles the sum will increase by at least 1/2, this is also only if we limit it to binary, in base 10 you get a more accurate sum approximation that is higher than the base 2 version. But either way, this shows that it does not tend towards any finite amount.

[@BlissBatch]

Why are you allowed to subtract ⅓ from one side, and then say it's less than the other side? Isn't ♾️-⅓ still ♾️? How can ♾️ be less than ♾️?

[@MyNameIsSalo]

1/3 isn't being subtracted... it's being replaced with 1/4 as 1/3 > 1/4. It simplies the problem and make a new series which is smaller than the origional series. If you can confirm that the smaller series does not converge, then the bigger one must not converge either. No one is subtracting from infinity, infinite is not a mathematical operator or value you can do math to.

[@457max]

I liked that integral proof.

[@MYNAME_ABC]

Even the sum of the reciprocals of ONLY the primes goes to infinity!

[@thexoxob9448]

This means the primes are dense over the natural numbers

[@Frozen_Hope]

The marker swapping feels like a magic trick, that's some skill

[@iankeith]

The second this clicked I laughed, it's a dang cool proof.

[@bpr214]

Integral of 1/n = ln(n) + c. Ln(n) goes to infinity as n goes to infinity. Therefore 1/n is not bounded. This can be generalized by saying any function n^-p for p equal to or less than one is unbounded. But is bounded for p greater than one.

[@kapsi]

Finally a math proof I can understand.

[@marianl8718]

1 > 1/2 1/2 + 1/3 > 1/2 1/4 + 1/5 + 1/6 > 1/2 1/7 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 > 1/2 Etc. Always the last term is the smallest, and their sum is greater than the number of terms multiplied by the last term.

[@connordelaney6254]

Holee this guy is the smartest man alive

[@dune22]

Great breakdowns!! Thank you.

[@JohnBerry-q1h]

boards are COOL !!