a spectacular solution to the Basel problem (sum of 1/n^2 via a complex integral)

[@Absilicon]

15:51, we learnt a lot about you 😂

[@jameswilson8270]

I think he was talking about "i" haha. that play on words is hilarious. btw, thanks for sharing bprp!

[@marbanak]

THAT'S AN IMAGINARY OBSERVATION.

[@WhattheHectogon]

As soon as I heard it, I knew the top comment was about it

[@nicholasleclerc1583]

marbanak I think it’s too *complex* of a message for us non-Rick-&-Morty fans to understand

[@tamirerez2547]

Very sharp thinking!! √-1 love this joke... (i love this joke)

[@firashadjtaieb6730]

I admire not the result but all the patience to write and explain every step :) if you want a problem with surprising result can you consider this one : take the polynomial (1+x+x^2)^n , note a_n the term of degree n , find an equivalent of it when n goes to infinity :)

[@blackpenredpen]

Thank you!

[@verainsardana]

i have done similar problem

[@Timorftw]

The best video on basel problem

[@cocoa1996]

Also, check out 3blue1brown's video on the same ;)

[@silverbladeii]

Existe uma prova muito legal no livro "tópicos de matemática elementar vol. 5" utilizando funções aritméticas.

[@ELBARTO2023-bg1rw]

this problem is always laughing in our faces with more than 100 proofs kwons but nobody until now know exactly why this sums really means, beacuse we can find the exact value of the zeta function on odd values, even its irrationality is not proved for zeta(2k+1) with k greater than one

[@DeeEm2K]

Please post more content which link two entirely different maths together like this man!

[@Alex-xc9sf]

15:51 “I don’t like to be on the bottom, I like to be on the top.” 😂

[@johnrodonis4186]

Hahaha!

[@Sam_on_YouTube]

That result is famous enough for there to be a proof wiki page on it with 7 proofs. This isn't one of them. You should add it to the list.

[@alxjones]

This isn't a proof, because the manipulations done with the series S aren't necessarily valid.

[@Sam_on_YouTube]

@@alxjones Thanks. You seem credible in what you said, in spite of the unfortunate man who shares your name. Sorry about that.

[@gregorykafanelis5093]

@@alxjones well if he proved the the series converges then everything is fine. The thing is that the series does indeed have a finite value so all the manipulations are valid

[@lukaskohldorfer1942]

@@gregorykafanelis5093 the problem is that in the complex case you don't always have ln(xy)=lx(x)+ln(y)... the complex logarithm isn't to be taken carelessly, so maybe already in the third line, there is a mistake.

[@gregorykafanelis5093]

@@lukaskohldorfer1942 well he then have to say that the ln is restricted for only 2π to 0 angles. But then we get down the rabbit hole. Point being, this proof is a long way from being mathematically strict but it is a nice way to calculate the value of the integral Also let's not forget the famous internet saying for mathematics If the result is correct then the method must be correct. This time we can turn our heads to the other side as you have to admit this proof us truly beautiful not mathematically strict, but certainly has some beauty

[@v_saaam]

Ok I admit, the end was really a surprise.

[@jibran8410]

Video was uploaded 5 hours ago...you commented 2 days ago.... Are you from the future

[@applealvin9167]

I’m from the future of the future

[@v_saaam]

I was from the past

[@xamzx9281]

that's an amazing integral! i will never stop learning from you

[@blackpenredpen]

xamzx thank you!

[@xamzx9281]

blackpenredpen btw can you integrate cosx/x from pi/2 to +inf like you integrated sinx/x from 0 to +inf

[@Achill101]

Well done. Your joy is infectious. And I had the joy of seeing yet another way on KZbin to calculate 1+1/2^2+1/3^2+1/4^2+... And your way is quick and doesn't require much higher math.

[@johnnath4137]

It was Jacques Hadamard (who proved the prime number theorem, along with de la Vallee Poussin) who said that the shortest path to a truth in the real domain often passes through the complex plane.

[@martinepstein9826]

You and Dr. Peyam are killing it with these elementary proofs of pi identities!

[@cliffhanger4930]

That was awesome, all kinds of mathematical ideas connected and I love how Euler famous e^(i*pi) + 1= 0 was utilized. Finish the proof with basic algebra. Great fun!

[@triniasta]

i think its funny to believe that under the |z|

[@pimcoenders-with-a-c1725]

You can also use parseval's theorem with f(x) = x, which also gives the solution of the basel problem! Doing the same with x^2, x^3, et cetera gives all the positive even solutions for the zeta function (Zeta(2), Zeta(4), Zeta(6) et cetera)

[@bernarddoherty4014]

Very very nice! Brings to mind the old saying..(going back to the 1600's actually) "there is more than one way to skin a cat." ....Google the phrase....basically saying there is always more than one way to arrive at the same result.

[@blackpenredpen]

Bernard Doherty I love this phrase!! Thank you. And I think that will be the perfect title of this video too! : )

[@blackpenredpen]

Bernard Doherty I will change the "skin" to "brush" so that the cat lovers won't go after me. : )

[@scathiebaby]

Thanks for that

[@pedjolinko]

One loophole is where you show that \int( ln (e^(ix)) (from 0 to pi/2)= i*pi^2/8. Keep in mind that: \int( ln (e^(ix)) = \int( ln (e^(i(x + 2*pi*n)) So, formally speaking, the result should be i*pi^2 * (1/8+ n) where n is an integer. Then you need to show that n must be 0.

[@vertechua]

Socrates shall rise once again. Hail! You foolish disrespect of a soul! For he might shower his generosity upon thy. Thou chains of burden shall burn, he shall return. Almighty shall lead you the way to freedom, and so you, by all means, shall accept a place in his bright side. Shall there lurk a soul who dares to challenge the lords incarnation shall not just suffer his own life, but for all forthcoming life forms of his. Thy must give a thought, indulging your conscience, and shall find it; not a reason but a need for acceptance. The absolute requirement of the knowledge shall put to prosecution all those who resist a change in the fallacy of satisfaction. You must one day bear the weight of responsibility, and so shall accompany the illusion of satisfaction. Socrates shall guide you to the moral, the only correct path, not just strickened with the fruits of knowledge, but shall show you the seeds of 'em. Let the world be free of the intoxication you have, shall I mention, gifted it in return of it's favour of the very air you thrive upon. Hail! For neither the mother nor shall the gods tolerate further of your actions, for in the most true sense, justice shall apply to all existing beings. This is not an ultimatum, rather I shall mention, is intended to be a forecast. Shall you be ready for the consequences, the mother shall never deprive her children of affection and forgiveness. Hail! Surrender, and beware, for you, now, are the only one...

[@MrNicolas609]

This is, in fact, one of the most beautiful videos i’ve ever seen

[@radiotv624]

Wow what a crazy cool Integral! Solving it seems fairly straightforward however where on earth did someone find out that this particular Integral leads to one of the most famous results in math? Either way, great video!

[@blackpenredpen]

Thank you : )!!!

[@rinostrozzino6467]

Wonderful and quite elementary way to solve the Basilea problem. Your videos rock! :)

[@blackpenredpen]

rino strozzino thank you!

[@ripansharma5259]

Wow!! What a crazy way to get to the solution of the Basel problem!! Respect Blackpenredpen👏👏

[@bouteilledargile]

I really like this proof of the Basel problem. I have just one hiccup with the technicalities: when you evaluated the power series for log(1+z) and integrated it, don't you have to prove its absolutely convergent? I understand that it would've been too technical but a mention would have been nice. Either way, great video!

[@martinepstein9826]

"don't you have to prove its absolutely convergent?" That would be a shame since the series is not absolutely convergent.

[@createyourownfuture3840]

@@martinepstein9826 Why? Is absolutely convergent different that just convergent?

[@martinepstein9826]

@@createyourownfuture3840 They're different. The simplest example of a series that's convergent but not _absolutely_ convergent is 1 - 1/2 + 1/3 - 1/4 + ... This converges to ln(2) but if you take the absolute value of each term you get 1 + 1/2 + 1/3 + 1/4 + ... which diverges.

[@createyourownfuture3840]

@@martinepstein9826 Oh...

[@MaximQuantum]

@@martinepstein9826 Is that what he meant with “absolutely covergent”?

[@krrishmaheshwari4860]

What a Brilliant way to prove this!!!❤️❤️ Proving one of the Best Equation in Maths in Best way!!!

[@johnnycrash1624]

Did you just find the craziest way to prove that (pi^2)/6 identity?

[@rot6015]

I KNOW RIGHT????

[@blackpenredpen]

Zvi did! : )

[@martinepstein9826]

What less crazy way of proving the identity do you have in mind? The infinite product for sinc(x) and 3Blue1Brown's lighthouses are both pretty crazy to me.

[@nicholasleclerc1583]

JohnnyCrash The simplest, should you say, ‘cuz the craziest thing we used was the complex (pun non-intended) definition of cosine in another problem, and everything was pretty straight-forward and self-explanatory....... compared to other proves

[@lukaskohldorfer1942]

​@@nicholasleclerc1583 the problem is that in the complex case you don't always have ln(xy)=lx(x)+ln(y)... the complex logarithm isn't to be taken carelessly, so maybe already in the third line, there is a mistake.

[@ManiFunctor]

Mind blown!

[@blackpenredpen]

Chris Hello : )

[@obedgarza6236]

You were so nervous trying not making mistakes. It was hilarious you were so excited i really like it. Congrats

[@blackpenredpen]

Obed Garza I was nervous trying to make sure I could fit everything on the board, as always : )

[@grindpalm]

Great job! You could have shown that S for even integers (1/2^2 + 1/4^2 + 1/6^2+...) = pi^2/24, which follows from S = S (odd) + S (even), i.e., pi^2/6 = pi^2/8 + pi^2/24. It's beautiful

[@xi-sca-38mainakchandra93]

Being a student of 11th grade, this is the only proof of the Basel Problem that is understandable for me. Thank you for the solution.

[@depressedguy9467]

Still in this proof it is required uniform convergence theorem and why we can manipulate that series

[@arequina]

Excellent video. Reminds me of complex analysis class I took in college. But that was so many years ago.

[@williammartin4416]

Thanks!

[@shashikumar7890]

No doubt 3brown video on this topic is simply awesome as it triggers the intuition behind the very answer. However, this video is nothing less than amazing for all who loves maths. In other words 3 brown Tries to answer in their every video, "why" certain things are the way it is. This video tells how you get there. I enjoyed both

[@ishakbasiru8561]

I really love mathematics. And I have been watching your videos and indeed, they have impacted my skills. I wish to meet you in life one day. Love your videos and hope to see more from you.

[@joshuacoppersmith]

What a fantastic surprise to start watching an integration video and end up with a great proof!

[@Koisheep]

Ok, but those who want to take the next step and make this into a proof may want to consider the following: 1. ln(e^z) is not always z in the complex plane. You have to prove ln(e^(ix))=ix for 0 < x < pi/2 (which I think is specific enough to be true) 2. You can't make 1+1/2²+1/3²+1/4²+...=(1+1/3²+...)+(1/2²+1/4²+...) unless you prove the series on the LHS converges first. I think ratio test doesn't work but maybe the Raabe criterion?

[@theoleblanc9761]

Yep I noticed it also, no proof of convergence. But it can easily be done if you assume that 1/1^2+1/3^2+...=π^2/8 (ie you have to justify a lot of things he did in the video before getting this result: ln(e^ix)=ix for x in [0;π/2], int of infinite sum is sum of infinite int...). 1/1^2+1/2^2+1/3^2+1/4^2.... 1/1^2+1/1^2+1/3^2+1/3^2+... Putting the terms like that, you see that every terms of Σ1/n^2 (n∈ℕ*) is less or equal than the corresponding term of the second series that obviously converges to 2*π^2/8=π^2/4 by hypothesis. So because they are series with positive terms, Σ1/n^2 (n∈ℕ*) converges. And the value is the one given in the video. In fact, without proving the convergence, he just show that IF it converges, then the value must be π^2/6

[@emmettweisz5880]

Yes! These steps are very important for the proof, it is incomplete without them. Good point! By the way, I think 1. can be proven easily by putting ln(e^(ix)) and ix in polar form, assuming 0

[@MarcoMate87]

I can at least justify the equation Ln (e^ix)=ix for x in [0;π/2]. We can use the formula ln(z) = ln |z| + i arg(z), which gives infinite results depending on which argument we choose for z. If we choose the principal argument of z Arg (z), with -π < Arg(z)

[@austinmitchell2652]

@2:19 Isn't the complex log multi-valued? Would that change anything?

[@azmah1999]

You restrict its definition so that it is single valued. The most common restriction is (-π, +π).

[@santiagoerroalvarez7955]

The path that the argument traces out doesn't cross a branch cut, so I guess we're OK

[@ashtonsmith1730]

@@azmah1999 i prefer from [0,2π]

[@mokouf3]

You solved basel problem with this amazing method? GREAT.

[@Eren_Yeager30]

Absolutely beautiful solution

[@edmundwoolliams1240]

This is fantastic. One of my favourite videos/proofs yet

[@DeeEm2K]

Amazing! I tried plotting the graph of that just to see whether the real part will be zero. I couldn't comprehend it lol

[@soupe2000]

I love this way of solving this problem

[@stefanoctaviansterea1266]

Amazing proof. I will show it to my teacher next year if he brings sum 1/n^2 up.

[@willyh.r.1216]

Very inspiring! People should love math by watching your video.

[@fiNitEarth]

OMG this is so cool! I'm so happy that I found this video 😍😰

[@Bodyknock]

Tangentially related to this video, 3Blue1Brown has a fantastic video called "Why is pi here? And why is it squared? A geometric answer to the Basel problem" which shows a geometric proof that 1/1^2 + 1/2^2 + ... = π^2 /6 using lighthouses around circular lakes. Highly recommend checking this video out (along with the rest of that channel, his videos are awesome! :) kzbin.info/www/bejne/ml7SZJh4btiZotU

[@BrotherSquid]

Doug Rosengard I honestly feel that 3Blue1Brown’s logic in that video is kinda dodgy. Don’t get me wrong though, I love his videos. Especially his essence of Calculus Series.

[@Bodyknock]

Just curious where you disagreed with his logic in that video. It all seemed pretty well laid out to me. Also he links to a paper in his description "Summing inverse squares by euclidean geometry" which was the basis of the video

[@shashikumar7890]

No doubt 3brown video on this topic is simply awesome as it triggers the intuition behind the very answer. However, this video is nothing less than amazing for all who loves maths. In other words 3 brown Tries to answer in their every video, "why" certain things are the way it is. This video tells how you get there. I enjoyed both

[@nickm1902]

This video has made me so happy :)

[@user-pw6qe7ur4q]

BEAUTIFUL!

[@garethxue8938]

Brilliant work

[@zwz.zdenek]

So good! So many twists and interesting approaches. I was expecting just some random formula as a result.

[@pspmaster2071]

Fascinating. Makes a lot of sense as you explain things. By the way I never thought of turning an integral into Macluaren or Taylor series if it is really difficult. Neat!

[@Aldron6]

then you can write the original integral as the integral of ln(2) + the integral of ln(cosx), to get that the integral of ln(cosx) is -pi/2 * ln(2)!

[@nischay4760]

So many results from this I am dying!😂

[@bouteilledargile]

yes that's actually a pretty famous integral as well

[@manishkumarsingh3082]

What is integral ln(cosx). ?

[@SanjeevKumar-js4mu]

Int 0 to pi/2 of ln(2) dx +Int 0 to pi/2 ln(sinx) =pi/2ln(2)-pi/2ln(2)) =0

[@gaunterodimm7219]

this gives the answer as 0

[@stevenp7991]

This is indeed spectacular. Thank you very much for this!!!!!

[@spartacus8875]

You are the great in mathematics....congratulations....

[@shardumachal]

The way he says super amazing..I am in just love with maths

[@mathmaths8380]

So much beauty in one formula

[@blackpenredpen]

YES!!!!

[@ianmi4i727]

This is Art. Math is Art. No matter the level!!

[@nathanisbored]

never knew you could prove it that way, wow

[@BrunoVisnadi1]

I was about to link you this video in Discord when I saw your comment. This proof was so beautiful.

[@unflexian]

Meeting you again, it seems? I guess it is a small world after all. Love your content!

[@gregoriousmaths266]

Which discord

[@titan4267]

thats genius lol . this is really surprising and i really like the way you explain it too. good job

[@anasettajani8123]

I wish you were available 10 years ago, you would have saved me from a lot of struggles

[@oscarfu9556]

This video deserves 1M views and likes❤️

[@moskthinks9801]

Zvi H. (or whomever told him) is the next Euler! Out of the random why is the time length sqrt(5)=2.236...?

[@blackpenredpen]

M. Shebl : )

[@coolzo2524]

Dang you worked really hard on this video good job 👍

[@matthewstevens340]

Personally I love the Taylor series approach using Sin(x)/x and treating it as a polynomial to solve the Basel problem, but this is amazing!

[@antoinemorrier9858]

That's the way Euler had an intuition about the result. But it is not rigourous enough ^^

[@linusschwan6299]

What an amazing ending! Great video!

[@mortezamodarres2470]

That was truely brilliant, wow

[@leviszhou711]

看到有的是通过给f(x)=x 傅立叶级数展开来给n平方分之一求和的。曹老师本次讲的这个过程也是很不错的！

[@semi8883]

that is actually insane. im blown away

[@___xyz___]

Holy moley that's a beautiful result

[@blackpenredpen]

Unknown Entity : )

[@matanfih]

Saw your videos ,BUT now you have a new subscriber

[@angelor_1354]

justo con esa idea comencé, pero tenia que saber el valor de la serie esa 1/1^2 + 1/3^2 + 1/5^2 ... etc. buen video, saludos desde Perú.

[@saksham1919]

how could one dislike a video so good

[@monika.alt197]

writing cos(x) as that exponential in the beginning, it looks like its equal to the hyperbolic cosine of ix, is there some relation between the two?

[@sunildey5887]

Learn a lot from you keep teaching

[@ShenghuiYang]

The process of derivation is a piece of art.

[@sumitprajapati821]

19:08 LOL 🤣🤣🤣 the funniest part!

[@manfredwitzany2233]

This can be calculated much easier: Just use the identity cos(x) = Re(e^(ix)) The real part operater can be brought in front of the integral due to linearity, which is just the easy left integral in your calculation. Then you calculate Re(i*pi²/8)=0

[@createyourownfuture5410]

Can you please elaborate?

[@AA-le9ls]

The integral operator is linear but not the logarithm function, so I don't think you can bring the real part operator anywhere in the intended integral expression.

[@Matthew-tu2jq]

This is pure magic 😍

[@Mathelite-ii4hd]

i saw the original proof for this with Euler's formula and i said,damn that is brilliant.but now after watching this,i can not even express how beautiful this proof is.

[@benoist13]

I really enjoy the way you're doing maths !

[@stephanbardel5681]

oh yeah, congratulations , a new way to prove this sum = PI²/6 ; very nice !!! Thx

[@aashsyed1277]

I love the way you solved 9r very nice 🙂🙂🙂👍👍👍 I think you will get more subscribers fast. Wish you best luck.

[@__-1234]

This is really brillant and simple...

[@sandyvu9652]

You're the best math teacher I've ever seen in my life it's soo amazing I love that🥲❤️‍🔥❤️‍🔥

[@hansulrichkeller303]

Schön gemacht - wie immer! Gratulation!!!

[@AngadSingh-bv7vn]

wow that was awesome how he was able to fit all the steps on the board

[@zmaj12321]

I have a question... can we plug in e^(-2ix) into the series of ln(1+x)? If x = pi/2, then e^(-2ix) = e^(-pi * i) = -1. Isn't that an invalid value for z? Yet it is part of the integral... Or do the bounds of integration not matter when evaluating using power series?

[@theinvisiblearmadilloofdea6204]

I have the same question.

[@rad858]

It's a good question... the short answer is that it's ok to plug these values in if they are limits. To make it more clear, you can write the upper limit as (π/2 - ε), do the integral, and then take the limit as ε → 0 afterwards.

[@easymathematik]

On the whiteboard he wanted to wrote | z |

[@ronald3836]

The whole thing is a bit shaky at |z|=1, so I would start by integrating log(1 + r * exp(2ix)) with 0 < r < 1. Work out what you get, check that it is continous in r, and check that you can now exchange taking the limit and summation because the coefficients go like 1/n^2.

[@JaydentheMathGuy]

Haven't seen integration in the complex world in so long (thanks a lot college professor) I almost forgot what e^i theta was. lol my precalculus teacher is getting so mad next time i visit her.

[@daigakunobaku273]

How can you integrate all these exponents over the region with logarithmic divergence?

[@Liesse_SportSante]

Very good video !

[@benattitude]

I love everything about this video

[@samuelromero6711]

Clever, but there’s a mistake. The power series for lnx only converges for values in the interval (-1,1]. Pi/2 is not in the interval of convergence.

[@patmark5886]

Best vidéo on the net!

[@1337ErkaN]

Best video on youtube.

[@andresxj1]

You, sir, fooled us! We thought we were dealing with an integral when we were actually looking for a series! FOOLER!

[@andresxj1]

Loved the video anyway ^_^

[@blackpenredpen]

: )

[@koenth2359]

Very brilliant and original solution to the Basel problem! Too many ad breaks though...😑

[@hankseda]

Hard not to be wowed by this video! One thing left to address is that the original integral is improper at the pi/2 end.

[@takyc7883]

Best explanation of eulers ever

[@inesantoniosanchezgutierre664]

Prettty, pretty amazing or awesome as you usually say. LIke we say in my pretty Nicaragua: you are a monster! You are a crack! Let me tell you that I follow you since the first time I found you in youtube. Have you thought in an analisis through any model for covid 19? I´ve seen a couple of them from Mathrock in Peru and Damian from Argentina. Man!!! you really love this beautiful science. Go on with that contagious enthusiasm!!!!!!!

[@marcelweber7813]

'Wow' is the right expression. That's just a thing of beauty. But I have one question: How does it fit with z not being -1 for using the fancy formula? You're integrating over an interval in which e^(-i pi x) will become -1 at pi/2.

[@frede1905]

I am not 100% sure about this, but I think that the things changed when we integrated the sum. Since z was e^(-2ix), we could see that when we integrated the sum, the terms were no longer on the form (-1)^(n-1)•z^n/n, but instead, it was on the form (-1)^(n-1)•z^n/(-2i•n^2), and therefore it had n^2 in the denominator, instead of n. Remember that the reason why we need to include the condition that abs(x)

[@marcelweber7813]

Sounds good for me, that's probably the reason why, but I'm still not convinced. It's right that z=-1 will make a diverging harmonic series, but isn't the main reason in this formula that ln 0 doesn't exist? So when we use the formula, we get a sum that is not defined for x=pi/2 which transforms back to z=-1. So if we were interested in finding a value for pi/2, we couldn't get an answer. This means we're integrating over a value that isn't defined at all. Could be infinity, could be 42. Maybe it can be made because this value is on the edge of our interval. I don't know. But just finding the anti-derivative and being happy that this one is defined on this interval doesn't connect it back to the original problem. For me it looks like integrating 1/x from -1 to 0, but not even knowing if 1/0 is the correct formula for x=0.

[@frede1905]

@@marcelweber7813 No, you are plugging in x =pi/2 AFTER you have integrated, and NOT before. Before you have done that, then sure, you can't plug in x=pi/2. But as I said, the fact that you are integrating changes everything, and now you CAN plug in x=pi/2. You have changed the sum entirely, and it is no longer on form of (-1)^(n-1)•z^n/n. It was the sum ON THAT FORM that was not defined for x=pi/2. But now the sum is on a different form, and NOW z=-1 is allowed (but it wasn't before). "Going back" to the original sum and noticing that your value for z (which is -1) is not allowed for that sum, doesn't mean that it is not allowed for the integrated sum. They are two separate sums with different properties. Even though that the second sum "came from" the first one, doesn't mean that they should have the same properties.

[@frede1905]

@@marcelweber7813 I understand your problem. When we assumed that ln(1+z) is equal to this infinite sum, then we already had to make sure that z is not -1. Then we integrated the sum, and suddenly we can plug in z=-1. But didn't we make sure that z is NOT -1 in the first place? Remember that the radius of convergence or whatever it's called changed during the integration, since now we can plug in z=-1 (the sum won't diverge). When we plug in x=pi/2 in the second sum, then we ONLY plug in x=pi/2 in the second sum. The second one is not connected to the first somehow, so that if you plug in something that is not legal in the first sum, then it is not legal in the second one either.

[@marcelweber7813]

Yeah you might be right. So after integration it's just fine to do so. But I still don't get the connection to our first definition... That's probably on me. I was thinking about an example, but I'm too stupid. What comes in my mind is maybe tan x. This one is defined on x=pi. Its integral -ln(cos x) is not (in R). So if we plugin pi into the function, does it still connect to the "definition" of this anti-derivative? Or maybe an even worse example: We say that all animals are cute exept for giraffes (z=-1) (everybody knows this). Now we take a look at the eyes of each animals (->integration) and find out, that the eyes of giraffes are cute as well. And now we say: fine, giraffes are cute too, because they have cute eyes. I know it's hilarious, but you see what I did there :)

[@draco23238]

This made my morning a "good morning"

[@yds6268]

Great proof! Very nice

[@Jenab7]

So in Σ(n=1,∞) n⁻² = π²/6 The odd terms contribute 75% of the total sum, while the even terms contribute only 25% of the total sum. Σ(n=1,∞) (2n−1)⁻² = 3π²/24 Σ(n=1,∞) (2n)⁻² = π²/24