a very nice approach to this limit.

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Комментарии : 71

@elidoz7449 8 дней назад

the way I did it is by Stirling's approximation, I justify it because it's like multiplying this limit by another limit which evaluates to 1, then I put them together into a single limit, which is allowed because both have definite values, and I simplified the factorial terms

@romain.guillaume 8 дней назад

I didn’t write it down but I have the exact same thought about Stirling approximation.

@nilshoppenstedt6073 7 дней назад

I had the same idea! 😅

@mirkorokyta9694 8 дней назад

To be able to multiply inequaluties you should ensure L-epsilon is positive not to lose the direction of inequaluties. In fact, in L is positive, one can always ensure this by choosing a proper epsilon, case L=0 must be discussed separately, and L cannot bez negative. This important reasoning is missing, which Is sad.

@TheEternalVortex42 8 дней назад

You could also just convert it back into the absolute value version before multiplying. Actually I think you can just stay with the absolute value version the whole time, there’s no specific reason to split it up.

@minamagdy4126 8 дней назад

It was already optimistic that you could prove this for negative L considering that, for even n, you can't even stay in the real numbers. For that, I imagine there is no avoiding complex analysis

@mirkorokyta9694 8 дней назад

@@minamagdy4126 a_n are supposed to be non-negative since you take nth root, therefore L cannot bez negative.

@michaelsheard4522 8 дней назад

Right. From the statement of the theorem L could be 0, which requires a slightly different proof.

@Czeckie 8 дней назад

since almost all terms in the sequence need to be positive you can just take the left hand side to be 0 instead of L-epsilon and it will work the same way.

@michaelsheard4522 8 дней назад

I believe the corollary to the Lemma is backwards. To be clear, the Ratio Test and the Root Test are both true, so in some sense each "implies" the other. But the Lemma allows a quick proof of the Ratio Test from the Root Test, not vice-versa. You will find it presented that way in some real analysis textbooks.

@samsonblack 8 дней назад

You are correct. I wrote a similar comment then found yours.

@samsonblack 8 дней назад

You've probably already noticed, but the natural corollary of your lemma is the *converse* of what you stated. In fact, any time you establish an implication between the hypotheses of two theorems with the same conclusion, you get an implication between the theorems in the opposite direction.

@samsonblack 8 дней назад

Details: abbreviating Theorems, Hypotheses, and Conclusions as Thm, H, and C, respectively, suppose: Thm₁ = (H₁ ⇒ C) and Thm₂ = (H₂ ⇒ C). Now suppose that you've established the implication H₁ ⇒ H₂ between the hypotheses. If you know that Thm₂ is true, then you can chain together the implications H₁ ⇒ H₂ ⇒ C, which shows that Thm₂ ⇒ Thm₁.

@holyshit922 8 дней назад

Stirling approximation gives 1/e

@anarcho.pacifist 5 дней назад

Nice! As an exercise for anyone reading, try to solve: limit(n->oo, (n!!)^(1/n)/sqrt(n))

@goodplacetostop2973 8 дней назад

12:38

@sonure6127 17 часов назад

You can also do it using reimann sum . The above expression can be shown to have the same value as exponential of integral of ln x from 0 to 1.

@vasilesinescu84 6 дней назад

This is what I knew as the Cauchy - d’Alembert criterion. It’s something equivalent to Cesaro-Stolz once you use exp logs. I see both results I cited above as l’Hopital for sequences - obvious as sequences are not differentiable functions. Enjoyed it, well done Michael😀

@andrey_kiev13 8 дней назад

I believe that you might imply the "root - ratio" limit equality as a following from the convergence radius of the power series definition, which is, precisely, this two formulas.

@dalitlegreenfuzzyman 6 дней назад

Stay encouraged, Michael. You are inspiring more people than you know.

@roberttelarket4934 8 дней назад

Can we get a backflip?

@hassanalihusseini1717 7 дней назад

No, we got 1/e.... not e.. 🙂

@roberttelarket4934 7 дней назад

@@hassanalihusseini1717: Very very very clever! Ha ha!

@alipourzand6499 8 дней назад

e is everywhere!

@danielevilone 4 дня назад

I'm an humble physicist, I see a limit with a factorial and just use the Stirling's approximation.

@seland2009 7 дней назад

In the proof of lemma, we multiple inequalities with (L-ε) on the left side, which requires (L-ε) to be non-negative to get true corollary inequality. Since a_n > 0, L cannot be negative, but it can be zero, in which case (L-ε) will *always* be negative. Thus, the case L=0 should be handled separately. To do that we can replace left hand side of the inequality with max(L-ε, 0) - it still holds since a_(n+1)/a_n is positive and also max(L-ε, 0) is non-negative allowing inequalities to be multiplied.

@wychan7574 7 дней назад

This problem appeared in uc Berkeley phd math qualifying examination. There are two approaches to finding the limit, this video gives one of it.

@othman31415 8 дней назад

Another way: Denote log the natural logarithm and u_n the sequence we want to calculate the limit we have log(u_n)=1/n*sigma(k=1..n){log(k/n)} and so the limit is just the integral of log(x) between 0 and 1 ( the Integra converges)

@mrmajestical 8 дней назад

The video was not made to find a quicker way but rather a beautiful and nice approach to the limit

@oryxisatthefront8338 8 дней назад

@@mrmajesticalit’s anything but beautiful

@padraiggluck2980 5 дней назад

Stirling approximation

@tolberthobson2610 8 дней назад

Assuming that for some constant, a_N, raised to the power of 1/n, as n tends toward infinity to be asymptotically approaching 1, wouldn't we have to demonstrate or presuppose a_N ISNT equal to zero? If it is equal to zero, more work would need to be done to ensure this statement holds true, right?

@11kravitzn 7 дней назад

The super sloppy method: If the value of the limit is x, this implies that n! ~ (n*x)^n Then, taking the ratio of the expression for n+1 and n, n+1~(n+1)*x*((n+1)/n)^n 1~x*((n+1)/n)^n Taking the limit, we find x=1/e

@xgx899 6 дней назад

This requires a proof that the limit exists. Once you add this proof, your approach becomes longer than the one presented.

@rainerzufall42 6 дней назад

Funny homework: lim_{n -> - \inf} (n!)^(1/n)/n = ... (to minus infinity, using Gamma function)

@rainerzufall42 6 дней назад

... = - 1 / e, what else?

@sergioc5 7 дней назад

The lemma can be proved with Cesaro limit much faster.

@TV-ll3kh 7 дней назад

In your thumbnail of this video, you placed the factorial sign outside the square root symbol.

@xinpingdonohoe3978 8 дней назад

Would you believe me if I guessed 1/e as the answer before watching? It came down to imagining it as (1/n)^(1/n) (2/n)^(1/n) ... (n/n)^(1/n), noting that each of those is

@dans.o.s.d.s6971 7 дней назад

what do you mean by (1/n)^(1/n).....?how did you reach that result?

@dans.o.s.d.s6971 7 дней назад

but i liked you idear , it's brilliant...when you used the logic of a positif converging sequence (and decreasing) (i'm used to maths in French, so some words may be awkward hhh.... i'm lazy to use translators..)

@xinpingdonohoe3978 7 дней назад

@@dans.o.s.d.s6971 (n!^(1/n))/n =(n!/n^n)^(1/n) =(1×2×...×n/n×n×...×n)^(1/n) =(1/n 2/n ... n/n)^(1/n) =(1/n)^(1/n) (2/n)^(1/n) ... (n/n)^(1/n)

@xinpingdonohoe3978 7 дней назад

It was from rearranging it and separating it into a load of terms, then grouping one bit from the numerator with one bit from the denominator. (n!)^(1/n) ×1/n =(n!)^(1/n) × (1/n^n)^(1/n) =(n!/n^n)^(1/n) =(1×2×...×n/n×n×...×n)^(1/n) =(1/n 2/n ... n/n)^(1/n) =(1/n)^(1/n) (2/n)^(1/n) ... (n/n)^(1/n)

@dans.o.s.d.s6971 7 дней назад

@@xinpingdonohoe3978 Oh, got it now....noice trick... with a little bit of reasoning, this quick guess works especially for eliminating wrong answers....

@arimermelstein9167 8 дней назад

Can someone explain how the limit being between L-ε and L+ε means that it has to equal L?

@SkorjOlafsen 8 дней назад

Definition of the limit. You have free choice of epsilon, so it's as small as you like. Informally, the value of the limit operator is the value as epsilon approaches 0.

@blackflan 8 дней назад

The inequality has to hold for every ε > 0. Intuitively, you are "sandwiching" the value of the limit inside an interval (L-ε, L+ε) with center L and width which gets smaller for every smaller ε you take, which is only possible if the limit itself is equal to L. For a more formal proof of that fact, you can try by contradiction, assuming that there exist another real number L' which is inside every interval (L-ε, L+ε) just as the limit, and that the limit is actually equal to L' and not L, but then you will find that after some value of ε (dependent on the difference between L and L'), L' will actually not be inside the interval, hence, the contradiction.

@arimermelstein9167 8 дней назад

@@SkorjOlafsen thank you

@arimermelstein9167 8 дней назад

@@blackflan thank you

@georgekh541 8 дней назад

Correct me if i'm wrong but i remember you doing the same video along time ago Edit: I just found the video ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8fI0S-HeYrQ.htmlsi=UZBtFMd77hPfsbwj

@sujitsivadanam 8 дней назад

I just looked at that video and yes, he does use the exact same method to solve this exact same limit.

@rockthemegaman2760 8 дней назад

Didn’t you technically use the converse of the lemma and not actually the lemma?

@ScytheCurie 8 дней назад

No, he used the lemma correctly in the "forward" direction. He showed that the limit of the ratio of consecutive terms equals 1/e. By the lemma, that implies that the limit of the nth root also equals 1/e, which is the limit we wanted to find. Admittedly, the way he worked it out on the board made it seem like he was using the converse because he started out writing the limit of the nth root and then ended by writing the limit of the ratio of consecutive terms. But in reality, he was using the fact that the limit of the ratio equals 1/e to imply that the limit of the nth root also equals 1/e.

@divisix024 8 дней назад

He used the lemma. He computed the limit of the ratio and concluded by the lemma the limit of the n-th root must be the same. He wrote it more compactly by writing everything in the same equation, but obviously you could’ve rewritten it as two equations, the first evaluating the limit of the ratio, and the second asserting the limit of the n-th root must be the value you got in the first equation.

@mirkorokyta9694 8 дней назад

@@divisix024 Exactly.

@divisix024 7 дней назад

@@ScytheCurie I mean, if you’re gonna directly use an if-then statement to calculate something, the thing you want to calculate would surely be the consequence (the thing in the then statement), which you use the precedent (the thing in the if statement) to compute.

@ScytheCurie 7 дней назад

@@divisix024 Yes, that's what I said in my original comment, Micheal Penn used the lemma correctly. That's also what you said in your original comment, so I believe we're in agreeance here?