a very nice approach to this limit.
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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
I didn’t write it down but I have the exact same thought about Stirling approximation.
I had the same idea! 😅
Correct me if i'm wrong but i remember you doing the same video along time ago
Edit: I just found the video
ua-cam.com/video/8fI0S-HeYrQ/v-deo.htmlsi=UZBtFMd77hPfsbwj
I just looked at that video and yes, he does use the exact same method to solve this exact same limit.
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.
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.
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
@@minamagdy4126 a_n are supposed to be non-negative since you take nth root, therefore L cannot bez negative.
Right. From the statement of the theorem L could be 0, which requires a slightly different proof.
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.
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.
You are correct. I wrote a similar comment then found yours.
No this is wrong as well as the video
Your statements are trivial
The corollary obviously should say:
"Ratio Test applicable implies Root Test applicable"
You find this in EVERY analysis textbook
If you apply your logic rigorously the corollary would state "Corollary: Ratio Test Theorem"
Big bs on this one
Stirling approximation gives 1/e
Can we get a backflip?
No, we got 1/e.... not e.. 🙂
@@hassanalihusseini1717: Very very very clever! Ha ha!
12:38
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.
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?
In your thumbnail of this video, you placed the factorial sign outside the square root symbol.
As (ln n + ln n-1 +...+ ln1)/(int_1^n lnx dx)=1 when n goes to infinity.
Let L be the limit. Then
ln L= 1/n * (int_1^n lnx dx) - ln n
=-1 - 1/n
L=1/e
I'm an humble physicist, I see a limit with a factorial and just use the Stirling's approximation.
e is everywhere!
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
This requires a proof that the limit exists. Once you add this proof, your approach becomes longer than the one presented.
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)
The video was not made to find a quicker way but rather a beautiful and nice approach to the limit
@@mrmajesticalit’s anything but beautiful
Funny homework: lim_{n -> - \inf} (n!)^(1/n)/n = ... (to minus infinity, using Gamma function)
... = - 1 / e, what else?
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.
Yeah, found it on real axis book and did with riemann sum
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😀
Can someone tell me what is wrong with my reasoning?
When we get the limit re-written as nth root of n!/n^n, inside the root we can expand that to be n(n-1)(n-2).../n*n*n*n...this can be split into fractions n/n * (n-1)/n * (n-2)/n * ....it can be seen that this entire expansion would converge to zero as n -> inf. But then if we introduce the encapsulating nth root term, any root of n < 1 pushes the term closer to 1, so I figured the final limit would just be 1.
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.
This problem appeared in uc Berkeley phd math qualifying examination. There are two approaches to finding the limit, this video gives one of it.
Nice! As an exercise for anyone reading, try to solve: limit(n->oo, (n!!)^(1/n)/sqrt(n))
Stay encouraged, Michael. You are inspiring more people than you know.
Stirling approximation
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.
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.
@@samsonblackno, thats not of interest. If it were the case, the corollary would just state "Corollary: Ratio Test Theorem"
But we know both are true, that would be trivial
The corollary SHOULD BE a short version for
"Ratio test applicable implies Root test applicable" and as we see should not only be denoted by the implication symbol
Edit: i actually watched the video and you are partially right
@@jgkgosjjdjd1382 Appreciate your enthusiasm, and I recognize that the directions of these arrows can get confusing (even Dr. Penn gets them mixed up once in a while), but if you read what I wrote closely, you'll see that I'm calling attention to this fact: The lemma shows that *if* the root test is true *then* the ratio test is true. This agrees with your assertion that "ratio test is applicable" (stronger hypothesis, weaker theorem) implies "root test is applicable" (weaker hypothesis, stronger theorem).
@@samsonblack no you still don't get it. The root test theorem is obviously true and the ratio theorem is therefore also obviously true. The corollary concerning "applicability" has practically nothing to do with the statements of the theorems. It says something about specific properties of given sequences.
The implication you still insist on being the natural has this crucial flaw that you refuse to see:
"Thm2 implies Thm1" itself as a statement would not imply "H1 implies H2"
You clearly understand that the word applicable is missing and you tackle this mistake by giving another statement that "misses" even more of the essential properties that the corollary wants to state
Edit: WOW i really never expected him to make this mistake. I'll be honest, i skipped this part of the video because it's such a basic corollary that i wouldn't write with the implication symbol. You are right about the video and how we get a proof of the Ratio test and not the Root test but i really didn't think that this would be worth mentioning so i still insist on what the natural corollary would be haha
Ridiculous discussion after all and to be precise your comment still stands wrong because we don't have an implication between the hypotheses of theorems
The lemma can be proved with Cesaro limit much faster.
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
what do you mean by (1/n)^(1/n).....?how did you reach that result?
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..)
@@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)
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)
@@xinpingdonohoe3978 Oh, got it now....noice trick... with a little bit of reasoning, this quick guess works especially for eliminating wrong answers....
Can someone explain how the limit being between L-ε and L+ε means that it has to equal L?
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.
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.
@@SkorjOlafsen thank you
@@blackflan thank you
Didn’t you technically use the converse of the lemma and not actually the lemma?
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.
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.
@@divisix024 Exactly.
@@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.
@@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?
I mean… ~ 1/n * n/e * (sqrt(2πn))^(1/n) = (1/e)*(sqrt(2πn)^1/n)
so lim = 1/e
I enjoyed this video. That lemma is so cool. 🧊