I did first year University mathematics in 1962. This brought back memories of my classes of nearly 60 years ago. We learned Raabe's test which, of course, we called "Raabies 'mad dog' test". As I recall there were also several generalisations of it that led to a sequence of more and more delicate tests, with comparison series of the form n^(-1), (n log n)^(-1), ... Am I confused about that? It was some time ago.
My favourite use of the Raabe convergence test is in the proof for the general binomial series when applied to real number exponents instead of just natural numbers. (1+x)^a = sum_{n=0}^{\inf} (a choose n) x^n for a in Real numbers. If a >0 on x in [-1,1], if -1 < a< 0 on x in (-1, 1], if a < -1 on x in (-1, 1). Raabe's test is needed for checking the boundary conditions on x=-1 and x=1 in the different cases when finding the radius of convergence of sum_{n=0}^{\inf} (a choose n) x^n.
Raabe's test, applied after the Ratio test, reminds me of repeated application of L'Hopital's rule until you get to a case that isn't one of the undefined forms of a limit. In both cases, once you get a definite answer, you can't expect to get any meaningful result by going to the next step. This then has me wondering if there might be an infinite series of such tests that one can apply until a result other than no-info is obtained. It is not, however, clear what the sequence of tests might be. Perhaps the next one is lim(n(1-n(1-a_(n+1)/a_n))) i.e. you subtract the non-committal limit expression from 1 and multiply by n and try again. We would need an example series for which Raabe's test yields no result to try this out...
Recursively, what you are looking for might look something like a test with lim(n->infinity)(b_i(n))=L , where b_0(n)=(a_[n+1]/a_[n]) and b_[i+1](n)=n*(1-b_i(n)) , and where you increment i to move to the next test (as well as switch whether L>1shows convergence or divergence, the reasoning for which is that you subtract something from 1 every time you iterate i).
Does there exist a Raabe's Test of second degree? I.e.: If the ratio test has L=1 and Raabe's Test has L=1 can you look at the expression lim n(1-n(1-a(n+1)/a(n)))? And if this is also equal to 1 can you extend this pattern infenitely often?
Never saw Raabs Test in my Calculus courses. Never heard of Bertrands Test or Gauss' Test. I will look them up. There is undoubtedly a whole slew of tests I haven't heard of. Man! There is always something new to discover and learn!
First time ever I see the Raabe's Test despite being in Three differents universities, two in Algeria and one in France.Very insightful as alwaysss thanks ! can't wait to watch tomorrow's marathon
@@maxwellsequation4887 yeah, in fact I started with a civil engineering curriculum then moved to common core curriculum major physics and chemistry and today math major I will be graduating by the end of the next academic year. It truly is an enriching experience
I didn't get at first what this test accomplishes and the sketch for the proof didn't clarify it either. The example, though, was quite useful. This test checks whether the ratio test approaches 1 slowly enough from below. with regards to the second most significative order of magnitude. And if so, then that means the series converges.
@@neilgerace355 no, as there is no "border series". There is no series such that everything that goes to zero faster will converge and everything that goes to zero slower will diverge. For example consider 1/n, now this diverges and but the p-series (1/n)^p for p>1 converges. So it seems like 1/n is the borderline case but it really is not, consider 1/(n*logn), 1/(n*logn*loglogn) etc. All those test like ratio, Raabe etc. are only utilising that we know about some convergent series, then we plug this series into comparsion test and create a nice formula out of it. What you learn in caluclus is just 1) integral test - gives a lot of examples of convergent series 2)comparsion test - allows us to compare to our known series. And this procedure is sometimes given a name like Raabe or ratio tets etc.
Homework for you boi 1) Sum from n = 1 to infinity (nπ Si(2nπ) + (1 - nπ^2)/2) Si is Sine integral function 2) Integral from 0 to infinity [x e^(x/2) - e^x +1]/[ x^2 e^(x/2) * (e^x -1) ]
a(n+1)/an < (n+1)^(-R)/n^(-R) a(n+1)/(n+1)^(-R) < an/n^(-R) By induction we can prove that: an/n^(-R) < a1 so an < a1 * n^(-R) since the series of term a1*n^(-R) converges, so the series of term (an) converges.
ok nvm i see its works but this explanation with binomial extension totally confused me (i proved it by using Bernoulli's inequality after few simple moves)
He may have a data limit on his Google account and doesn’t want spam images to take up space. It’s easy to load an image to Imgur then copy the url into Michael’s form.
Wolfram alpha let me do about 3000 terms before quitting on me (or asking for a pro upgrade) and I had to write the numerator as product with Pi notation. Not sure if there’s a more efficient way but it wasn’t recognizing triple factorial notation. The partial sums were indeed getting to about 2.6.
I never saw this test before! It's awesome, but out of curiosity I searched for more information on Wikipedia and a really famous math website of my country (it's called Youmath if you are interested, but it is written in italian tho) and found a different form for this test: the ratio had the terms of the sequence swapped and 1 was the quantity subtracted, not the other way around like in the video. Can somebody explain me the difference, please?
@@HS-fw2ed If you refer to Bernoulli's inequality, unfortunately here the exponent is negative. You have to Taylor expand or use MVT to prove that here.
I wonder what weird test you would have to use to implicitly compare with [(log n)^p n]^-1 or whatever. Or [(loglog n)^p log(n) n]^(-1), and so on. I'm sure someone's figured something out, though this family isn't a theoretical worst possible rate of convergence, either.
Just as an aside to this, many, if not most students of series think that divergent series are bad and can't be used for anything. One thing is that they think divergent series becomes infinite. This isn't true, and actually divergent series are more useful than convergent series, and often converge to the correct answer faster than an equivalent convergent series. There is an excellent series of UA-cam videos by Carl Bender at ua-cam.com/play/PL43B1963F261E6E47.html. The first handful of the videos are very understandable. As it goes on it becomes a lot harder since the entire series is for graduate level physics student at the Perimeter Institute for Theoretical Physics, but still very interesting.
I would like to suggest a problem. My name is Isam Ahammed from India. My question is the sum of all the integers which are less than N and co-prime to it is 1/a*N*phi(N), a belongs to positive integers. Find a.
a=2. The trick is to observe that whenever (m,n)=1, then (n-m,n)=1 too. So, you can rewrite your sum's terms as n-m for m with (m,n)=1 and add the the sums (like Gauss did for the sum of consecutive integers).
I think all these tests are worthless, since the amount of calculation you put in there is basically the same as in calculating the asymptotic behaviour for large n. So for instance if asymptotically a_n = f(n) + o(f(n)) (small o), and the convergence/divergence behaviour of f(n) is known, then you can always deduce the behaviour for a_n.
interesting, we did learn about this
Same, but not in the form he wrote
Same.
Different
I did first year University mathematics in 1962. This brought back memories of my classes of nearly 60 years ago. We learned Raabe's test which, of course, we called "Raabies 'mad dog' test". As I recall there were also several generalisations of it that led to a sequence of more and more delicate tests, with comparison series of the form n^(-1), (n log n)^(-1), ... Am I confused about that? It was some time ago.
My favourite use of the Raabe convergence test is in the proof for the general binomial series when applied to real number exponents instead of just natural numbers.
(1+x)^a = sum_{n=0}^{\inf} (a choose n) x^n for a in Real numbers. If a >0 on x in [-1,1], if -1 < a< 0 on x in (-1, 1], if a < -1 on x in (-1, 1).
Raabe's test is needed for checking the boundary conditions on x=-1 and x=1 in the different cases when finding the radius of convergence of sum_{n=0}^{\inf} (a choose n) x^n.
Raabe's test, applied after the Ratio test, reminds me of repeated application of L'Hopital's rule until you get to a case that isn't one of the undefined forms of a limit. In both cases, once you get a definite answer, you can't expect to get any meaningful result by going to the next step. This then has me wondering if there might be an infinite series of such tests that one can apply until a result other than no-info is obtained. It is not, however, clear what the sequence of tests might be. Perhaps the next one is lim(n(1-n(1-a_(n+1)/a_n))) i.e. you subtract the non-committal limit expression from 1 and multiply by n and try again.
We would need an example series for which Raabe's test yields no result to try this out...
Recursively, what you are looking for might look something like a test with
lim(n->infinity)(b_i(n))=L
, where
b_0(n)=(a_[n+1]/a_[n])
and
b_[i+1](n)=n*(1-b_i(n))
, and where you increment i to move to the next test (as well as switch whether L>1shows convergence or divergence, the reasoning for which is that you subtract something from 1 every time you iterate i).
@@minamagdy4126 where can i find more information about this
Does there exist a Raabe's Test of second degree? I.e.: If the ratio test has L=1 and Raabe's Test has L=1 can you look at the expression lim n(1-n(1-a(n+1)/a(n)))? And if this is also equal to 1 can you extend this pattern infenitely often?
There are Gauss's test and Bertrand's test and other more higher order tests of similar type to Raabe's.
Only for those sequences which satisfy the difference equation.
a(n+1)/a(n)=n/(n+1)
Same thoughts.
I was wondering the same thing.
Fred
Never saw Raabs Test in my Calculus courses. Never heard of Bertrands Test or Gauss' Test. I will look them up. There is undoubtedly a whole slew of tests I haven't heard of. Man! There is always something new to discover and learn!
In fact this was the only way to solve the series I was asked in my final Calculus 1 exam
Kummer's test covers it and it basically implies many other useful tests like Gauss's test and Bertrand's test.
This is what I have wanted!!!!
First time ever I see the Raabe's Test despite being in Three differents universities, two in Algeria and one in France.Very insightful as alwaysss thanks ! can't wait to watch tomorrow's marathon
U studied in 3 Universities
Awesome
@@maxwellsequation4887 yeah, in fact I started with a civil engineering curriculum then moved to common core curriculum major physics and chemistry and today math major I will be graduating by the end of the next academic year. It truly is an enriching experience
I didn't get at first what this test accomplishes and the sketch for the proof didn't clarify it either.
The example, though, was quite useful. This test checks whether the ratio test approaches 1 slowly enough from below. with regards to the second most significative order of magnitude. And if so, then that means the series converges.
Nice example and work through. Thank you for sharing.
Epsilon-delta proofs are the coolest in calculus. Looking forward to that live stream!
They were the Bane of my existence on analysis
In*
U think they are cool? Wow!
What a maniac!
6:28 This product can be expressed with Γ function 3^(n+1)Γ(4/3+n)/Γ(1/3) (there was problem to name this product)
If the Ratio Test and Raabe's Test fail (L=1), is there yet another test we can use?
For example Gauss test. Almost all test are constructed/derived by comparsion to some convergent series, so you can build better and better test.
@@lukashorak9639 Cool. Then, is there a test that will never fail?
@@neilgerace355 no, as there is no "border series". There is no series such that everything that goes to zero faster will converge and everything that goes to zero slower will diverge. For example consider 1/n, now this diverges and but the p-series (1/n)^p for p>1 converges. So it seems like 1/n is the borderline case but it really is not, consider 1/(n*logn), 1/(n*logn*loglogn) etc. All those test like ratio, Raabe etc. are only utilising that we know about some convergent series, then we plug this series into comparsion test and create a nice formula out of it. What you learn in caluclus is just 1) integral test - gives a lot of examples of convergent series 2)comparsion test - allows us to compare to our known series. And this procedure is sometimes given a name like Raabe or ratio tets etc.
You might say that Cauchy test is one that never fails, but thats basically just rewritten definition of convergence.
@@lukashorak9639 Thanks! But if there is no boundary between 1/n and 1/n^2 how the sum (from the video) n^-R could be definetely convergent?
In the Russia we did. Also we did learn Еrmakov's test (de.wikipedia.org/wiki/Kriterium_von_Ermakoff) and many others.
I think I did learn this in my calculus 101 year. The reason it's not used much is for most cases, it offers nothing new compared to ratio test.
11:01 Good Place To Stop
No homework today, see you tomorrow for the live stream!
See ya.
I wonder, though, why we have never seen Michael smile
@@reshmikuntichandra4535it would be a Good Place To Smile
Homework for you boi
1)
Sum from n = 1 to infinity
(nπ Si(2nπ) + (1 - nπ^2)/2)
Si is Sine integral function
2)
Integral from 0 to infinity
[x e^(x/2) - e^x +1]/[ x^2 e^(x/2) * (e^x -1) ]
We did cover it, that's why I'm here:)
How does the ratio being less than the ratio of a convergent p-series imply convergence?
By the ratio test.
a(n+1)/an < (n+1)^(-R)/n^(-R)
a(n+1)/(n+1)^(-R) < an/n^(-R)
By induction we can prove that:
an/n^(-R) < a1
so an < a1 * n^(-R)
since the series of term a1*n^(-R) converges, so the series of term (an) converges.
a_n is bounded by 1/3^n then we may apply the convergence given by geometric series.
no idea what happens at 4:26 Why this inequality (1-R/n)
ok nvm i see its works but this explanation with binomial extension totally confused me (i proved it by using Bernoulli's inequality after few simple moves)
Penn Sir, pls allow to add images in the form for suggesting questions
He may have a data limit on his Google account and doesn’t want spam images to take up space. It’s easy to load an image to Imgur then copy the url into Michael’s form.
Now im curious what this limmit equals to
About 2.6 probably
Wolfram alpha let me do about 3000 terms before quitting on me (or asking for a pro upgrade) and I had to write the numerator as product with Pi notation. Not sure if there’s a more efficient way but it wasn’t recognizing triple factorial notation. The partial sums were indeed getting to about 2.6.
First test is called the d’Alembert criteria in France :)
In Greece too.
I never saw this test before! It's awesome, but out of curiosity I searched for more information on Wikipedia and a really famous math website of my country (it's called Youmath if you are interested, but it is written in italian tho) and found a different form for this test: the ratio had the terms of the sequence swapped and 1 was the quantity subtracted, not the other way around like in the video. Can somebody explain me the difference, please?
If you multiply the expression in the video by a_n/a_(n+1), this doesn't change the limit (because the limit of a_n/a_(n+1) is assumed to be 1)
3:45 I didn't get it how did u proof that R will be lesser than that equation instead of greater than ?????
Good Place To Start At 0:01
I (with time) understood all of this but @ 11:01. How/why is 1-R/N less/equal to the binomial expansion?
I think you mean 4:30. If that is the case, then it is because of: (1+x)^m≥1+mx
@@HS-fw2ed If you refer to Bernoulli's inequality, unfortunately here the exponent is negative. You have to Taylor expand or use MVT to prove that here.
I wonder what weird test you would have to use to implicitly compare with [(log n)^p n]^-1 or whatever. Or [(loglog n)^p log(n) n]^(-1), and so on. I'm sure someone's figured something out, though this family isn't a theoretical worst possible rate of convergence, either.
Yea we didn't learn it but it was our homework problem to prove it
Why is it not taught
I did!
Just as an aside to this, many, if not most students of series think that divergent series are bad and can't be used for anything. One thing is that they think divergent series becomes infinite. This isn't true, and actually divergent series are more useful than convergent series, and often converge to the correct answer faster than an equivalent convergent series.
There is an excellent series of UA-cam videos by Carl Bender at ua-cam.com/play/PL43B1963F261E6E47.html. The first handful of the videos are very understandable. As it goes on it becomes a lot harder since the entire series is for graduate level physics student at the Perimeter Institute for Theoretical Physics, but still very interesting.
If my recollection is correct, it's Raabe Duhamel ratio test.
I would like to suggest a problem. My name is Isam Ahammed from India. My question is the sum of all the integers which are less than N and co-prime to it is 1/a*N*phi(N), a belongs to positive integers. Find a.
a=2. The trick is to observe that whenever (m,n)=1, then (n-m,n)=1 too. So, you can rewrite your sum's terms as n-m for m with (m,n)=1 and add the the sums (like Gauss did for the sum of consecutive integers).
I think all these tests are worthless, since the amount of calculation you put in there is basically the same as in calculating the asymptotic behaviour for large n. So for instance if asymptotically a_n = f(n) + o(f(n)) (small o), and the convergence/divergence behaviour of f(n) is known, then you can always deduce the behaviour for a_n.
I think mst black/red pen will like this video 😊
1:29 1:33 1:38
Isn't it just l'Hôpital's rule?
NEED TO LEARN
IN BIBLE ?
stats:365likes-3650views
🤩🤩✔
In Russia we actually study it.
We learn it in Russia.
Raabes' test looks like Rabies test lmao