Real Analysis | Monotone sequence theorem example 2.

[@xriccardo1831]

Professor Penn is the best youtube professor in my opinion. His arguments are coincise and he ALWAYS proves everything. Even the little lemmas.

[@DarkCloud7]

A coin's size is concise. Nice unintended (?) play on words there. Also I agree with you.

[@flux4162]

Michael : 'messes up' Also michael : 'deep breath ' 'Looks at the camera' 'Corrects himself Continues .... Utter legend, keep it the good work, love from the uk👏🤘

[@hoodedR]

Lol yeah

[@VerSalieri]

I love that... the unlocking of techniques in Calculus 2... like a game where you need more experience to reach the next level....and just as ironic....you unlock the powerful tools exactly at the moment when you can survive without them. One note, if the sequence is decreasing, you can just prove it has a lower bound. And if it’s increasing, you can just prove it is bounded above. Forgot to say.... you are a legend.

[@garrycotton7094]

6:02 - 😂 That reaction was priceless.

[@goodplacetostop2973]

15:56

[@mkskms322]

Damn bro im laughing.

[@thesecondderivative8967]

5:08 I don't think we have to cheat. 3n/(2n+1) is less than 3n/2n which is equal to 1.5 since we're making the denominator smaller. Comparing far right and far left, we have the required inequality. 13:49, Is it fine to prove that it's decreasing by induction?

[@westxlcr]

Thanks for the explanation. My mind got stuck in the mud during this part of Real Analysis. Your examples cleared it up for me!

[@filipchris245]

At 7:15 it has to be 3/2(2n+1) not 1/2(2n+1).Nothing changes but is good to be correct.Everything else was great.Very nice video,thanks.

[@jeffersonazevedo]

Traz, posteriormente, uma série sobre funções de variáveis complexas. Parabéns pelo canal.

[@golvanontheroad]

Nice video, very educational. Just wanted to point out that in this case one could easily give an explicit formula for the sequence, and then convergence is obvious.

[@backyard282]

0:50 you said that we cannot calculate the limit yet without the epsilon delta, but in a previous video of real analysis you have proven several limit properties that we can apply here. First we can divide numerator and denominator by n, then use the limit of quotient, sum, constant, and finally that the limit as n goes to inf of 1/n is equal to zero.

[@MK-13337]

Did you listen to the whole quote? "At this point in this type of course..." If you are taking elementary real analysis and you are doing these kind of convergence proofs you aren't gonna have all these tools yet. His videos aren't a university course. At the course they will ask you to prove this limit with the epsilon N (or epsilon delta, same thing basically) method.

[@rustemtehmezov9494]

Hi, can you make about CMC (cyberspace math competition) questions which shared nowadays

[@jideofororamah4399]

Very helpful video

[@jiaweisun8890]

minus 2 on both sides we have a_{n+1}-2=1/2(a_n-2). then the sequence a_n-2 is geometric. We then can solve the general term for a_n-2, and then the general term for a_n= 2+2^{2-n} Thus lim a_n=2

[@1346bat]

Love this channel....germany

[@jeffersonazevedo]

Even today in Brazil, the memories of the 7 x 1 in the world cup continue. Whenever your country has a breakthrough, we joke 7 x 1 with our lack of governance.

[@yossefswelam265]

another approach an = 4 - 1 - 1/2 - 1/4 - ... - 1/2^(n-1) and this is a geometric series taking the limit as n goes to infinity the result will be 4 - 2 = 2

[@anuragjuyal7614]

I also do the same thing . Just plugged a_n to get equality between a_n-1 and a_n+1 and the repeated till i got to similar result

[@psy7669]

an=3n/(2n+1) < 3n/2n = 3/2

[@kazebaret]

I would hope that there is a value, let's say A, that will never be reached by the series. A> a_n => A>3n/(2n+1). Rearranging: n(3-2A) > A. But n, A>0, thus 3-2A>0, thus A

[@jonaskoelker]

I argued approximately the same: If a_n < A for all n then 3n/(2n+1) < A so 3n < A*(2n+1) = 2*A*n + A implying (3-2A)*n < A [I think you flipped the ordering here] in turn implying either 3-2A = 0 or n < A/(3-2A) for all n. The latter is impossible by the Archimedean principle, so 3-2A = 0 implying 3 = 2A implying A = 3/2

[@tgx3529]

The sequence a_n+1=0,5 a_n +1 is decreasing, y1=0,5 x+1, y2=x, y12, if limit exists, so L=0,5 L+1, so L=2, my sequence is decreasing & not negative, so the limit realy exists.

[@andresfeliperengifojaramil4782]

Very usefull video, thanks, 👍

[@ostdog9385]

I like your channel more than bprp...

[@electroskylightgaming4085]

Traitor

[@ostdog9385]

@Adam Romanov I never said bprp isnt good... but i agree, flam has gotten worse

[@electroskylightgaming4085]

@Adam Romanov really? I dont know man....I think he has one of the best videos on special functions like Gamma and Polygamma.Apart from that, he is a pretty nice guy,even if some people might not like his somewhat 'inappropriate' jokes.But either way,I think he is a decent math youtuber.I enjoy him

[@alainbrizard4719]

If a_1 = b and a_n+1 = a_n/2 + 1, then a_n = 2 + (2 b - 4)/2^n. So lim a_n = 2 for all b.

[@NotYourAverageNothing]

Or suppose that 3n/(2n+1) > 3/2 Assuming 2n+1 > 0, 6n > 6n+3 → 0 > 3 false, so 2n+1 < 0 → n < -1/2, which is also false.

[@sentipy8990]

Who the hell is that serial disliker who puts 1 dislike under each video? xD

[@yoavshati]

Probably a bot

[@derekfordyce9]

Cracking the Cryptic has that problem too.

[@sairamganna350]

In place of that use (an+1-2) =.5*(an-2) So an+1-2 = . 5^n * (a1-2) Lim n->inf an+1 - 2 = 0 or an = 2

[@sumitprajapati821]

At 5:10 I think it should be 1

[@cacostaangulo]

Just transform the series to lim(n->inf) of (1+2^(n-1))/(2^(n-2))

[@dileep_j203]

👍👍👍👍

[@shohamsen8986]

Take the limit in the defn, then you get l=l/2+1. Thus l=2

[@mikeburns]

Nailed it!

[@shohamsen8986]

@@mikeburns unfortunately, this is assuming the limit exists and is unique.

[@mikeburns]

@@shohamsen8986 Not saying this is how to _solve_ it. This is a quick way to set the bound before the proof by induction (rather than trying some values and proceeding with a guess and check)

[@shohamsen8986]

@@mikeburns "This is a quick way to set the bound before the proof by induction". But I am not setting any bounds, l is the limit of a_n. I think i may have misunderstood your statement.

[@michamazur2179]

When we show it is bounded can't we just take first term, and since an is increasing just take lim n->inf an = 3/2? Way faster and easier.

[@kendakgifbancuher2047]

The point of video is to show that bounded = convergent if monotone on some examples. Video is not about taking limits

[@michamazur2179]

@@kendakgifbancuher2047 yeah i know, but i mean isn't it good way of showing that an is bounded. Because still the rule applies, but I prove that an is bounded other way.

[@kendakgifbancuher2047]

well, sometimes when you are dealing with basics, you have to restrict yourself in "knowledge" to avoid circular thinking (A is true because B is true, which is true because A is true). For example you cant use Lhopital's rule (I hope I spell it right) to prove anything. Because taking derivative requires to use limits and not the other way around. In case of this video its always useful to keep things simple without any additional knowledge to avoid possibly trap yourself into circular reasoning.

[@kendakgifbancuher2047]

yet again, if you want to *solve* something, you can use any valid means you want

[@michamazur2179]

@@kendakgifbancuher2047 okay, now i understand. Thanks