you forgot the a=pi on the top so you need to divide by pi. =1/24-1/(8*pi)

Yes!!! Please derive the inverse hyperbolic trig function sum result on your second channel!!!

You should set *both* a and b to pi, then you avoid the problems raised by other commenters. Final answer is 1/24 - 1/(8*pi), which matches a numerical approximation.

The final answer you get is negative whereas all the intermediate series have strictly positive terms.

at b=0, b/sinh²(bn) will blow up to ∞, hence the final answer is coming out to be negative whereas the sum is strictly positive

As written in other comments, the hyperbolic sine identity at the end of the video Min 8:00) only holds when ab=pi^2, where a,b are both positive Here ere is a reference Bruce Berndt, Ramanujan notebooks Vol. II, p. 245, equation (1.16).

Fantastic derivation! One of the things I love about this channel is its directness. Even little mistakes are valuable, because they force me to review more carefully the part of the problem that I couldn't follow. Then I will often look in the comments, to see if others saw the same thing. I think you should bless your favorite "fix" in the comment section to an error in the video to make sure other people can find it too. I love how you keep me solving problems every day, and introducing me to corners of maths i never ran into before.

b=0 is not allowed. b=a=\pi is the correct choice

9:07 - All the constituents are positive, so it makes no sense that you got a negative result. There's got to be a mistake somewhere.

i don't see why the second tool even needs the terms with b. you could just set a=b and divide by 2, so that you have sum(a/sinh²(an))=a/6-1/2

The hyperbolic sine identity at the end of the video only holds when ab=pi^2, where a,b are complex numbers with positive real parts.

Really cool technique with the partial derivatives! The only thing I will say is in the final step shouldn't we use a = b = pi to derive the final result? I don't think using a = 0 is allowed since sinh(0) = 0? And if you do use a = b = pi you get the same answer as what wolfram alpha says: 1 / 24 - 1 / (8*pi).

The result from Wolframalpha differs: (pi - 3)/(24 pi)

I think there is a problem, because at the end there was no "a" in the numerator, it means that we should divide first part of the solution by pi. Another problem is an indeterminate form (when b = 0), in limit form there is [0/0] form by L'H for at least 3 derivatives, and saying that "x has the same behavior as sinhx near 0" is also incorrect, because we having squared term. Of course we can reduce power of the sinh in denominator, but we were forgot anything of this in this video. There are some problems at the end of the solution. But also, this problem is an interesting one, and with problems above it increases curiosity of finding correct solution. Thanks for this video, also it's interesting to look for deriving another Ramanujan tool.

Wrong Final answer is (pi - 3) / 24 pi

there was a mistake at the end

A friend pointed this out to me: There is a problem with the identity introduced at the 8 minute mark. Assume a and b are positive and a+b < 6. Then the right hand side is negative, but the left hand side is composed of positive terms, so it must be positive (!). So, at minimum, it appears some restrictions are needed for the identity, restrictions that are not stated here.

I love your videos and I enjoy them so much, thanks

Sweetness of blatantly unchecked and wrong answer!

Your video is amazing and it really makes sense to summing technics.

you can't set b=0, because sinh^2(0)=0 .and b/sinh^2(b)is like 1/b i.e. 1/0. you can set a=b=pi and then we will get : 1/24-1/(8*pi)

That last Ramanujan identity looks very wrong. It seems as though you could split the sum into the sum with a only and the same sum with b only. Then you get f(a) + f(b) However, besides being redundant, it looks like near zero the sum should diverge b/sinh²(bn) ~ 1/bn² for n >1/b ==> these terms become insignificant and fewer as b→0, and the also add contributions with the same sign as the rest so they would even cancel out anything. The sum should go like 1/b * zeta(2), but the RHS suggests it approaches -1/2. How?

I'd love to see that hyperbolic sin equation derived, it looks crazy!

Love your shirt! Phoebe Bridgers is great.

I would be interested in seeing the derivation of the sinh identity.

This is the coffee compliment I needed to start my Friday!

SO COOL!

In order to be able to interchange the bounds of summation at 3:45, is it enough to check that the original sum converges? Or more specific requirements are also needed to ensure that the interchange is valid?

It diverges faster if you include i in the exponent

but B can not be equal to 0 because it would give you a 0/0 cause sinh(0) = 0

Well, we have the Weierstrass function which is an infinite series which is nowhere differentiable, so shouldn't we establish that the infinite series that we are doing is indeed differentiable before moving the differential operator around?

Fascinating video and I would've loved it, but I'm actually kind of annoyed that you didn't derive the sinh identity because it's incredibly unclear and the main gap between the problem and the solution.

Something must've gone wrong. $\frac{\pi}{24} - \frac14 < 0$, but you're only summing positive terms?!

At several intermediate steps I thought you were going in another direction with re-interpreting an expression as a sum or integral of something. Is linking the original sum to this black box reciprocal-hyperbolic-sine-squared identity that merits its own video really the most direct route to final answer?

another crazy approach would be to multiply and divide by 2$\pi$ and then expand the internal sum using bernouli numbers

I would like to see a proof of the second formula in 'use'.

This can be verified easily by using the well-known Ramanujan sum which is f(0)/2 +the infinite sum from 1 to inf of f(n) = i times the integral from o to inf of (f(ix) - f(ix))/(exp(2*pi*x) -1).

Awesome method! What is the rigorous justification for using partial derivatives within discrete expressions of kind like in the video?

lovely

I can't believe the final answer because it is negative. There must be an error somewhere.

Isnt that a negative number? This cant be correct

Hey, Michael! Would you please derive the 2nd tool? Thank you.

Yes please! More Ramanujan magic on second channel would be great :)

Always sanity check your results, if you start with a positive number the result should not be negative. Also pi in the exponent of the original sum could be arbitrary number ... nice sum though ...

Olvidaste demostrar porque esa suma con sinh

Isn't b=0 make the second fraction undefined? sinh(0) = 0 for what i know

3:05 The series converges absolutely, so we may change the order of summation, immediately getting the result 6:12 without partial derivatives. You sum over *n* first and simplify with the help of the generalized geometric series: *∑_{k = 0}^∞ \binom{k + m}{m} * q^k = 1 / (1 - q)^{m + 1}, m ∈ ℕ_0, | q | < 1*

The final answer is negative??

Waiting for the follow up of the unproven identity that surely doesn't make sense as it stands if you set b=-a.

Am I wrong or you missed a π?

how you can take derivative with index variable m

Seria bueno demostrar la segunda identidad utilizada, la cual no es simple

2:37 Sum of a product equals product of a sum? Explanation perhaps?

probar la parte 2 de use

That identity cannot possibly be correct. Suppose a=b=1, then the left-hand side is clearly positive, but the right-hand side gives -2/3.

There is a mind blowing appearance of it in the "Black body radiation"

Pi does not seem to make any contribution in this sum. It just serves as a constant. You can take it out to save some writing time.

You should at least explain why b=0 works, since sinh(0)=0 and you divide by it. I know L'Hopital handles it, but someone may get confused by it.

the answer is a negative number!? Instead, you should set a = b = pi, which would give -1/2 instead of -1 and so forth

Bruh i feel like my mathematical abilities have gone down studying for high school

There must be a little error somewhere: Mathematica gives the result 1/24-1/(8π) for this sum.

Unfortunately your answer is wrong as many others have already pointed out, so you should fix it.

Wanna see the proof of the sinh identity pls 💪🏻

this is wrong again! 7:37 this point and beyond is wrong the answer is 1/24 - 1/(8 π) also, the infinite sum of [n/(e^(2πn)-1)] is the same as the infinite sum of [25^-n+(1-8π)^-n] starting at n=1.

It’s not real

asnwer=1n ! isit

Sir Srinivas Ramanujan 🥳

Dear Michael Penn, please correct the answer in the video. Sinh(0) is 0, I do not believe in dividing by 0 sir. Set b to pi and do the math. Thank you for the video!