Correct me if I'm wrong, but I think the mistake is writing sum( |a_nb_n| ) = lim |sum( a_n-a_{n+1} ) sum(b_k)| This is because the second term is actualy |sum(a_nb_n)| where the modulus is outside the summation, not inside!
@@vishithvas5626 the third equality on the right is nonsense: on the left you have a sum of the absolute values, and on the right the absolute value of a sum
Okay - I may be wrong and apologies if so but my take is: there was no editing error.. Michael was being good teacher and emphasizing difference by parts aiming for a(0) >= zero so he let it be performed in front of us more that once
The first proof is not correct because, for instance, by taking a_n = 1/log(n) we have a monotone decreasing sequence converging to 0, and by taking b_n = (-1)^n / n we get a bounded series (the "alternating signs harmonic series" is convergent), however the sum of the absolute values has general term |a_n b_n| = 1/(n log(n)), so it's divegent. Which means the claimed bound is not true, and it's not surprising at all: absolute convergence is a tad too much. It can be fixed by taking the norm of the difference (a_1 b_1 + ... + a_n b_n) - (a_1 b_1 + ... + a_m b_m), which turns out to be bounded by M(a_n - a_m), concluding that the sequence of the sums a_n b_n is Cauchy, hence convergent, since a_n is Cauchy. Also, the entire thing can be generalised to complex numbers or Banach spaces by assuming the series of |a_(n+1) - a_n| is convergent and a_n tends to zero.
@@violintegralyup u are right. He mentions testing absolute convergence, but he doesn't the equality for the absolute value. It is definitely troublesome
16:11 I calculated this sum with generating functons and complex numbers Generating function for sine , G(t) = Im(1/(1-exp(i*x)*t)) Generating function for partial sum S(t) = 1/(1-t)*G(t)
@@droid-droidssonIt doesn't need to converge. It just needs to be bounded. And showing geometric series are bounded for |q|=1 is simple, you have formula for each partial sum.
@@kokainum it has to be uniformly bounded, there has to be one fixed circle around the origin that none of the partial sums ever leave. That is not the case for q=1, so your argument cannot work for any q such that |q|=1.
They did learn me this test during my Calculus courses. And, as an extra, I had to figure diverging sequences when one of those conditions was not fulfilled. That time, it was quite tricky.
Funnily enough, we _did_ learn about this in calculus, but that was the maths portion of a physics degree, where applications of this convergence test are quite common,
the transfomation you made at first it's called in french as " la transformation d'Abel"
English textbook ruined!
Must use french edition!
"La transformation d'Abel"? What the hell is that!?
@@PPolycephalum if you don't know what i am talking you could have just ask
@@achrafsaadali7459No, it was a joke. Just trying to make someone laugh :)
(See "Homer Simpson le grill")
You can also say Abel's transformation in English
@@stefanalecu9532Abel transform
Summation by parts (analogue with integration by parts). This should be better known. Very useful in a lot of places.
There's and editing error, but it's a good thing for us to find where's the error in the first proof attempt :D
Correct me if I'm wrong, but I think the mistake is writing sum( |a_nb_n| ) = lim |sum( a_n-a_{n+1} ) sum(b_k)|
This is because the second term is actualy |sum(a_nb_n)| where the modulus is outside the summation, not inside!
Video editing error? Anyway, the second time Michael goes through the proof is crisper.
The first proof is wrong
@@Nolord_Why though??
@@vishithvas5626Because sum of |a_n b_n| isn't bounded when a_n = 1/n and b_n = (-1)^n
@@vishithvas5626 the third equality on the right is nonsense: on the left you have a sum of the absolute values, and on the right the absolute value of a sum
Okay - I may be wrong and apologies if so but my take is: there was no editing error.. Michael was being good teacher and emphasizing difference by parts aiming for a(0) >= zero so he let it be performed in front of us more that once
The first proof is not correct because, for instance, by taking a_n = 1/log(n) we have a monotone decreasing sequence converging to 0, and by taking b_n = (-1)^n / n we get a bounded series (the "alternating signs harmonic series" is convergent), however the sum of the absolute values has general term |a_n b_n| = 1/(n log(n)), so it's divegent. Which means the claimed bound is not true, and it's not surprising at all: absolute convergence is a tad too much.
It can be fixed by taking the norm of the difference (a_1 b_1 + ... + a_n b_n) - (a_1 b_1 + ... + a_m b_m), which turns out to be bounded by M(a_n - a_m), concluding that the sequence of the sums a_n b_n is Cauchy, hence convergent, since a_n is Cauchy.
Also, the entire thing can be generalised to complex numbers or Banach spaces by assuming the series of |a_(n+1) - a_n| is convergent and a_n tends to zero.
Can someone explain the equality at 5:16? I thought it was supposed to be the absolute value of the whole sum of anbn, not the sum of /anbn/.
He made a mistake there and for some reason he didn't edit it out
@@violintegralyup u are right. He mentions testing absolute convergence, but he doesn't the equality for the absolute value. It is definitely troublesome
And indeed, sum a_n b_n does in general *not* converge absolutely. Just try a_n = 1/n and b_n = (-1)^n
16:11 I calculated this sum with generating functons and complex numbers
Generating function for sine , G(t) = Im(1/(1-exp(i*x)*t))
Generating function for partial sum S(t) = 1/(1-t)*G(t)
7:19 Wtf? This is so wrong...
edit : The second proof is correct though. He forgot to remove the first one.
I think using sin n as imaginary part of exp(i*n) and showing geometric series with q equal to exp(i) is bounded is way simpler.
|exp(i)|=1, therefore the geometric series cannot be shown to converge or diverge by that method.
@@droid-droidssonIt doesn't need to converge. It just needs to be bounded. And showing geometric series are bounded for |q|=1 is simple, you have formula for each partial sum.
@@kokainum it has to be uniformly bounded, there has to be one fixed circle around the origin that none of the partial sums ever leave. That is not the case for q=1, so your argument cannot work for any q such that |q|=1.
@@droid-droidsson For q=1 it's not the case but for every other q such that |q|=1 it very much is the case.
I've got a dejà vu experience mid video...
They did learn me this test during my Calculus courses. And, as an extra, I had to figure diverging sequences when one of those conditions was not fulfilled. That time, it was quite tricky.
Beautiful. Wasn't aware of that, thanks 🙂
The third condition means than the series of (b_n) is bounded. I think it's easier to state and understand this way personally
Funnily enough, we _did_ learn about this in calculus, but that was the maths portion of a physics degree, where applications of this convergence test are quite common,
Hi,
Thank you very much, I think you answered a question I asked a few videos ago.
9:46 : good place to start.
20:43
We did learn that in Calculus.
Same.
Seems like there was a video editing issue with this one?
Seems like he has no editor any more, and he hardly has the time to do this job properly himself.
There are very good book : T. J. BROMWICH , introduction to infinite series. 1908.
😊
we did learn that cryterium on calculus but im a phisics major so maybe thats why (i havent used yet)
I remember that in calculus
I did learn it. It was in Calculus 3
This is covered in calculus 3 and I used it in calculus 2
Not in Calc 3 in the US. Series are done in Calc 2 but this test is done in a real analysis course.
Good new photo profile 🗿👍
The second proof is much more comprehensible
The title of the video where you proved summation by parts formula is "The sum of n/2^n two more ways!"
Oh my god new profile picture 😮😮
Yes , very nice, I like it too.
We did in Russia)
There is something odd here. It seems he shows absolute convergence which is a nonsense .
We have two nearly identical first thirds, don't we?
The first proof is wrong, he probably forgot to edit it out.
A simple question. Why dont u pronounce dirichlet as u pronounce chloe or echo. Fourier is not four year. Etc... Just saying maikel.
You can spell out "you."
@@forcelifeforce why owe you?
First
Ew... Also no you weren't.