Proof: harmonic series diverges | Series | AP Calculus BC | Khan Academy

[@frnchdazzled]

"I think you see whats going on here, THIS IS EXCITING." = best part of my day

[@VCT3333]

The real deeper mystery of the harmonic series is not that it diverges, but how slowly it diverges. Even to get to 100, you need to use 10 exp(43) terms. That's only to get to 100. Imagine getting to a billion, and realizing billion is basically zero compared to infinity!!

[@tomasbeltran04050]

everything is basically 0 compared to infinity

[@OpPhilo03]

i did not belive that harmonic series diverge.

[@carlmoller807]

This is one of the coolest things I have seen in math so far. Mind blown.

[@brendan7224]

That was indeed exciting when he said “I think you see what’s going on here”. Great video!

[@sukheth]

u just changed the way I see math !!!!! awesome !!!!! good going.

[@urmishharibhakti6493]

Damn. This is why math is beautiful.

[@davoohm2779]

What I find amazing is that a series like this one diverges while the sum of all inverse square numbers (1 + 1/4 + 1/9 + 1/16 + ...) converges to a relatively small number (pi^2/6) even when both 1/n and 1/n^2 both approach 0 as n increases.

[@rileyjeffries1282]

And it’s also crazy that the inverse of that sum (6/pi^2) is the probability that two random positive integers share no common factors

[@crocoddile]

it's very interesting, at the first time I see the harmonic series I thought that it should must be converges lol, but now I know how to proof it diverge. thank you!

[@EthanMah-i4v]

excellent explanation ! ive watched many videos on harmonic series and you explained it best

[@thetruth5084]

This will forever seem like pseudo-mathematics even though I know it to be true.

[@gordongorgy9148]

It's a lot more intuitive if you use the integral test

[@GrayYeonWannabe]

@@gordongorgy9148 really? i think this is much more intuitive than using the integral test

[@TheFarmanimalfriend]

From my perspective, I know it to be false. I have a degree in biochemistry and was going over some of the calculus I learned for my degree. It is false. Read my answer above.

[@GamerTheTurtle]

@@TheFarmanimalfriend Your biochemistry degree < Engineering Degree therefore your biochemistry degree < Mathematics Degree by the comparison test therefor your "answer" is false

[@erichuang5971]

@@GamerTheTurtle haha he was joking chill bro

[@maticjurac]

I just heard this in calculus class today, I just heard that harmonic series are divergent, took my dog out for a walk earlier, absolutely livid at how someone could even come to such a seemingly illogical conclusion, and then I somehow thought of this exact proof like 2/3rds down and was astonished.

[@williamgomes216]

My brain just exploded. That was so cool!

[@Fazil_az]

Thank you very much sir...This helped me for my end semester calculus preparation...

[@shabbir.akhtar]

Finally a good explanation of it!!

[@tonychen3306]

Sal if u pick certain set of numbers in original harmonic series u'll actuall find that those numbers add up equal to 1 and there are unlimited number of that kind of sets in this series, so the harmonic series is actually divergent

[@AtlasReburdened]

That's a beautiful notion too. Divergence by means of containing infinitely many possible convergent sets. That's a much more fascinating way to view countable infinity than just "all the rational numbers", even though they're equivalent statements.

[@zacklight]

Actually any positive integral's reciprocal other than 1 should work, but this really helps~~

[@nicholashalden601]

This proof was made before calculus it's the great thing about it, simple and elegant :D

[@wizardcandy2449]

1:07 is my favorite part

[@sorooshb007]

Great video thanks a lot

[@apocalipto91]

Elegantly beautiful !!!

[@frozenstrawbs]

stewart calc presents this series on the same page as the test for divergence (i.e. limit of the sequence not 0) without noting the paradox. wtf

[@cheofusi3562]

That is just.....wonderful

[@MohamedNidabdella]

There is another way to go around which is finding that two sub-series converge to two different limits, it's more simple than Nicole's proof, but yeah it's good to have two technics rather than one

[@sijiahuang6936]

elegant!

[@mwabatikambenji4995]

ooor one half 5:06 lol .This is awesome

[@gasun1274]

actually this isn't obvious at all why this works. that's what real analysis is for.

[@santinodemaria2818]

Where u talking about Nicole or Tartalia at 1:06?

[@VVarun97]

Great, Now I just have to learn what is convergence and divergence 😂

[@CaribSurfKing1]

Yes, it diverges, however as n increases its affect on the total has less and less affect ( plot the graph of n vs total ), which could be classed as convergent

[@peace4aIl]

Carib thats wrong.

[@CaribSurfKing1]

***** As n increases the total increases at an increasingly slower rate

[@peace4aIl]

CaribSurfKing1 Ok. Some apples are green. What's the point?

[@MC-mx1mt]

a sequence that has the limit 0 does not necessarily means it is convergent. However, a convergent series must have its sequence have a limit of 0.

[@JakeRoberts-o2q]

Amazing

[@solidwaterslayer]

y don't schools teach the way you do

[@danielgoodman3578]

On "largest power of 1/2 that is less than or equal to ", we can compute that (1/2)^(log(5)/log(2))=1/5 when looking at a replacement value for 1/5. So this approach then only considers...I'm not sure of the right word here, but for lack of a better one, "whole number values" (e.g., (1/8)^(1/3) as opposed to messy in between things like the logarithmic values above?

[@Letsgettouchy]

Finding common roots for the fractions w/ denominators that aren't powers of 2 is unnecessary since your proving divergence with the direct comparison test. You only need to show that your comparison function is smaller than the function getting analyzed and the smaller function diverges

[@hanskywalker1246]

Great

[@epb111]

I understand how this proof works. What I don't understand is how if the sequence 1/n converges to 0 then its series would not also converge. If the sequence is eventually 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 ,0 . How is the corresponding series not convergent. But then you look at this proof and it is contradictory to that and makes just as much sense. I'm so confused.

[@matthewjimenez4720]

You will never actually be adding 0s. it will be extremely close, but you will forever be adding tiny fractions that will add up to 1/2 as proved in this video.

[@sakjdh]

It si also possible to take the indefinite integral of 1 over n and then find out that the area is equal to infinity which by integral test mean that series diverges

[@TheFarmanimalfriend]

The harmonic series diverges because it does not converge to zero, not because it is incorrectly summed to infinity. Nicole Oresme was incorrect in his summation.

[@Letsgettouchy]

Then what does it converge to, tough guy? (Hint: its a number that is infinetly big!)

[@OpPhilo03]

1/n^2 series also diverges? same concept use.

[@NuEM78]

I think I just had a braingasm.

[@FirstLast-ws7zw]

Beauty.

[@tomasbeltran04050]

2^ceil(log(x)/log(2))

[@MrKool-uj6rk]

but then at one point youll have a group of terms that must add up to 1/2, thats infinitely getting larger.

[@MrKool-uj6rk]

basically what im trying to say is, the amount of terms needed to successfully create a grouping must be finite to equate to 1/2 but the amount of terms needed approaches infinity, meaning at one point you will need an infinite amount of terms for a grouping to equate to (1/2) which cant make sense because the sum will never be (1/2) rather something infinitely close to 1/2. infinity am i right

[@thestunyz9566]

Approaches infinity, **but never equals it**. It could take the largest number known to mankind to create another 1/2, but, the series goes on infinitely. And infinity is always greater than non-infinity.

[@muralik98]

How to group terms... how many terms are allowed in a group

[@harrytaylor4360]

shouldn't it be the *least* power of 1/2? If I look for the largest power of 1/2 that is less than one, I can just keep getting bigger powers --> smaller than 1. Am I missing something?

[@ezioauditore6963]

the greatest exponent, not the whole thing

[@AhmedKhashabaa]

OMG That mind blowing

[@matthewjames7513]

I wonder what the value of S divided by the harmonic series approaches? I'd guess 0, but I can't prove it. Help plz :D

[@pianoforte17xx48]

Sorry for the late reply, but generalizing the harmonic series, it's S1 of (1/n). Generalizing the smaller series, it's S2 of (1+[n/2]). Finding the limit of S2/S1 will approach infinity, thus this ratio is divergent as well.

[@quriositysquared1028]

I finally get this

[@jimmyhsp]

does every interval (of the root/1) appear in the harmonic series?

[@anoirtrabelsi8645]

I worship Sal !

[@GustavTang]

I fucking love you man! you helped SO much!!! thank you.

[@amblypygi6651]

My mind is blown

[@iaaan1245]

wait... can you use integral test to proof that the area of the function of 1/x is finite, and thus the harmonic series converges as well....?

[@tintun8918]

Area under 1/x is not finite. It is logrithmic. f(x)=1/x .... F(x)=ln(x)+C. And as x-> infinity, ln(x) goes to infinity.

[@sangamesh2727]

[@miguelriesco466]

Well and where is the proof that each power of one half appears that many times?

[@LOVE..Sherelle]

I know that that's what it is, but it's just not making sense in my head.

[@nohandleid]

this proof is so elegant yet simple that you could probably teach it to a monkey

[@tommcgill8601]

I was thinking, but how do you prove that that we will be provided by a sufficient amount of powers of 1/2 between each number to add up to 1/2. And I did a little direct proof myself for this, We can see it intuitively with the counting numbers by subtracting consecutive powers of 2. eg.. 4-2 = 2 8-4 = 4 16-8 = 8 however this isn't a proof either. Although we can prove here that this pattern always holds: that is 2^n - 2^(n-1) = 2^(n-1), you can get from one side of this equation to the other using simple index manipulation. So yeah there will always be a sufficient amount of numbers in between to perfectly make another 1/2.

[@shawn7117]

Is there a reason why this will happen?

[@lexisanchez8711]

How does 1/3 become 1/4

[@thelosttomato4020]

It doesn't become 1/4, the guy is trying to make another series, each of whose terms is less than or equal to the terms of the harmonic progression. If he can prove that that diverges, then the harmonic series diverges. The closest power of 1/2 that is less than or equal to 1/3 is 1/4.

[@adairpowers1696]

PLEASE MAKE A MUSIC THEORY VIDEO !!!! :(

[@Migrond]

This seems strange to me. Basically what is happening is that instead of adding each of the ever decreasing terms to a sum (a sum which should be going nowhere because the terms are of no relative value) you are taking infinitely many terms, adding them together and having them represent the next term. Which is essentially cheating because its circumventing the rate which n goes to infinity! As n approaches infinity, the series S has to step infinitely far ahead to collect all the terms it needs to produce another 1/2-term. Essentially what I'm saying is that each *actual* term in S has been multiplied by a number approaching infinity to make this work. Comments/thoughs?

[@robert-andreidamian790]

+Nate H this is viable because infinity has no boundry. Infinity + Infinity is equal to 2 * infinity, but that's also just 1 infinity. I recommend you to watch a video about the hilbert's hotel paradox. It would explain this to you better than I ever could.

[@Migrond]

+Damian Robert I'm familiar with Hilbert's Hotel, but I'm not sure it addresses the problem I'm having. Hilbert's Hotel deals with the idea of infinity, while this is about the rate which an infinite sum gets there. Although as I write this, I realize maybe a sum is always considered to diverge - provided that each term is a positive, regardless of how relatively small the value is, and how 'slowly' the sum goes to infinity. Meaning my earlier problem didn't have to do with the concept of infinity, but rather the idea that the infinite sum of n, and 1/n both are considered to 'diverge'. Even if 1/n will probably never get beyond a sum of 30, given a number of terms that exceeds all practical application.

[@CaffeCraft]

Heym, hope I'm not too late on this, but your statement about any sum with all positive terms diverging is false! A great counter-example to it is the infinite sum whose generic term is 1/(2^{n}) (famous series used in Zeno's Paradox), or, in fact, any infinite sum whose generic term is 1/(p^{n}), for p > 1. The first series converges to 1 (this is proved by noting that 1 is an upper bound of any parcial sum of that series, and that 1 is the smallest of upper bounds, thus, by definition of convergence, it can only converge to 1).

[@CaffeCraft]

It's viable because of how we define "divergence" and "convergence". A series converges to a real number c if (in practical non-formal terms) you can show me that, for any tiny neighborhood of c (for example (c-0.00001, c+0.00001)), you can gather up a finite number of terms of your series so that it is in that small neighborhood. If you can prove that given any tiny neighborhood of c, there is an amount of numbers of your series that sums up to something close to c, then it converges to c! And a series diverges if it doesn't converge. Does that help you understand how the harmonic series fails (by comparison to S) to achieve that property of convergence to any real number?

[@billylee5624]

I know it's cheating, but have you ever heard of the geometric series proof? You also manipulate the series to get the rule.

[@hellowill]

Idk but I still look at it and doesnt seem like it obviously diverges lol

[@johngddr5288]

its not obvious, but it does!

[@jirihvozda4096]

It is obvious because of "necessary condition of series" {a}_n is a series, n going to infinity, then a_n is going to zero, otherwise series diverges

[@johngddr5288]

@pyropulse clearly not until you learn it.

[@anwarulbashirshuaib5673]

I get that it diverges, but why? I just cannot convince myself. Let's say that nth term is 1/n, and (n+1)th term is 1/(n+1), so the limiting value of (1/(n+1))/(1/n) as n->infinity =1(according to convergent test formula), so the series must converge. But here you showed that this series diverges! How on earth!?

[@pianoforte17xx48]

The ratio test says that if the limit is 1, then the test is inconclusive, since the ratio that's multiplied by each term is neither increasing nor decreasing it. That's why you should head to other tests. One of them is the integral test. It's gonna be ln(n), so its limit at infinity is infinity, then it's divergent. Think of the integral as the sum of areas, thus being a series. To convince your self more, don't get tricked by the fact that since the infinite term of the series is 0 means that the series is convergent. It's the addition of every number between 1 and 0, and between them is an infinite set of numbers, so it should diverge. I hope that helps!

[@jaieeke]

s=3/4

[@jasonyao3753]

8 people failed their math exams

[@KrishnaKumar-if9mg]

holy smokes

[@KENSON1997]

I am not a good mathematician but I do have a query : 1+2+3+4+5+6+7+.......∞ =-1/12 the S in your video is S= 1+1/2 +1/2+1/2+1/2+1/2+1/2+1/2+....... , cant it be interpreted as S=1+(1/2+1/2+1/2+1/2)+(+1/2+1/2+1/2+1/2+1/2+1/2)+........ which can be rewritten as S = 1+2+3+........∞ then can the starting sum ( 1+1/2+1/3+1/4...........) still be diverging ?

[@KENSON1997]

Thiago Ferreira Not convincing as in my starting sum 1+2+3+4+5+6+7+.......∞ =-1/12 or my 'conclusion' :S=1+(1/2+1/2+1/2+1/2)+(+1/2+1/2+1/2+1/2+1/2+1/2)+........ = S = 1+2+3+.......∞ ?

[@thiagorisingforce]

Kenson Li both, infinity is a very hard concept to grasp, these sums are not growing at the same pace, so when you equal infinities always seems like a leap of faith. But that is just me.

[@juliusdictatorperpetuus1205]

With divergent series like 1+2+3+... which go to infinity, you cannot really add up certain terms within the sum, as it can create complications. So, 1+1/2+1/2+... is not equal to 1+2+3+... . Another way to see it is that 1+2+3+... approaches infinity 'faster' than 1+1/2+1/2+..., as the partial sums increase at a faster rate, so the 2 series are not equal.I hope I answered your question with that.

[@KENSON1997]

Thiago Ferreira I see how it wont work now , thanks

[@MohamedNidabdella]

***** it does actually, using the zeta function, we can prove that it's -1/12. but this is applied to String theory, and does not mathematically make sens.

[@MG30001]

kinda boring and I don't get it. I expected sH to be always be 1,99999... but sS is larger than 2.

[@strummerbaum4409]

This totally explains the useless description that Stewart Calculus gave for this proof 🙏🙏🙏

[@chrisstuart6651]

How is the description explained?