Real Analysis | Monotone sequence theorem example 2.

We give two examples showing sequences converge using the monotone sequence theorem.

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[xriccardo]

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

[flux]

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👏🤘

[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.

[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!

[filipchris]

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.

[thesecondderivative]

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?

[garrycotton]

6:02 - 😂

That reaction was priceless.

[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.

[backyard]

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.

[rustemtehmezov]

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

[jiaweisun]

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

[tgx]

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

[yossefswelam]

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

[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<3/2.

[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.

[alainbrizard]

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.

[cacostaangulo]

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

[sairamganna]

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

[michamazur]

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.