There is a nice trick to calculate this limit.

Подписаться 301 тыс.

Просмотров 64 тыс.

50% 1

Suggest a problem: forms.gle/ea7Pw7HcKePGB4my5
Please Subscribe: ru-vid.com...
Patreon: / michaelpennmath
Merch: teespring.com/stores/michael-...
Personal Website: www.michael-penn.net
Randolph College Math: www.randolphcollege.edu/mathem...
Randolph College Math and Science on Facebook: / randolph.science
Research Gate profile: www.researchgate.net/profile/...
Google Scholar profile: scholar.google.com/citations?...
If you are going to use an ad-blocker, considering using brave and tipping me BAT!
brave.com/sdp793
Buy textbooks here and help me out: amzn.to/31Bj9ye
Buy an amazon gift card and help me out: amzn.to/2PComAf
Books I like:
Sacred Mathematics: Japanese Temple Geometry: amzn.to/2ZIadH9
Electricity and Magnetism for Mathematicians: amzn.to/2H8ePzL
Abstract Algebra:
Judson(online): abstract.ups.edu/
Judson(print): amzn.to/2Xg92wD
Dummit and Foote: amzn.to/2zYOrok
Gallian: amzn.to/2zg4YEo
Artin: amzn.to/2LQ8l7C
Differential Forms:
Bachman: amzn.to/2z9wljH
Number Theory:
Crisman(online): math.gordon.edu/ntic/
Strayer: amzn.to/3bXwLah
Andrews: amzn.to/2zWlOZ0
Analysis:
Abbot: amzn.to/3cwYtuF
How to think about Analysis: amzn.to/2AIhwVm
Calculus:
OpenStax(online): openstax.org/subjects/math
OpenStax Vol 1: amzn.to/2zlreN8
OpenStax Vol 2: amzn.to/2TtwoxH
OpenStax Vol 3: amzn.to/3bPJ3Bn
My Filming Equipment:
Camera: amzn.to/3kx2JzE
Lense: amzn.to/2PFxPXA
Audio Recorder: amzn.to/2XLzkaZ
Microphones: amzn.to/3fJED0T
Lights: amzn.to/2XHxRT0
White Chalk: amzn.to/3ipu3Oh
Color Chalk: amzn.to/2XL6eIJ

Опубликовано:

27 дек 2021

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 144

@christopherallen3353 2 года назад

Beautiful solution. Not sure I would have figured out how to re-engineer it on my own.

@alokdhardubey5030 2 года назад

Wow 😯, you proved Stirling's approximation in between like it was nothing. I had to do a lot of work in proving it in my second semester

@lunstee 2 года назад

My immediate thought on seeing the naked n in the denominator is to try to bring it in closer to the factorial so we can let the large n's cancel each other out. So: (n!)^(1/n) / n = (n!)^(1/n) / (n^n)^(1/n) = (n!/n^n)^(1/n) From here, we can pair up factors in the factorial with factors of n^n n!/n^n = n/n * (n-1)/n * (n-2)/n .... * 1/n 1 * (1-1/n) * (1-2/n)...*1/n) This is clearly between 0 and 1 since all the individual factors are, and taking the n'th root of this also remains between 0 and 1. Being between 0 and 1, we can take the logarithm, letting us work with a sum of terms instead of product of factors: log( (n!/n^n )^1/n) = 1/n * [ log(1/n) + log(2/n) + log(3/n)...+log(1-1/n) + log(1) ] The limit of the sum as n goes to infinity is the integral: integral(x=0..1) log(x) dx = x*log(x) - x [for x 0,1] = -1 As this is the logarithm of the expression given, the expression itself is e^-1 or 1/e

@goodplacetostop2973 2 года назад

16:59

@criskity 2 года назад

I thought to myself "It has to be less than 1. And it probably has something to do with e. So I'm guessing that the answer is 1/e." Not exactly a rigorous proof! But my guess was right!

@brian554xx 2 года назад

I call this mathematical intuition, and it helps more often than it should.

@qovro 2 года назад

@@brian554xx Can you prove that intuition helps more often than it should, or is that just your intuition?

@brian554xx 2 года назад

@@qovro Perfect query.

@user-lh5hl4sv8z 2 года назад

@@qovro lmao

@mr.soundguy968 2 года назад

So you didn't prefer 1 over e

@Robert_H. 2 года назад

Stirling formula: n! = 2 * pi * n * (n/e)^(n) (exact for n -> infty) (n!)^(1/n)/n = (2 * pi * n)^(1/n) * n/e * 1/n = (2 * pi)^(1/n) * n^(1/n) * 1/e Limes n towards infinity leads to 1 * 1 * 1/e = 1/e.

@uy-ge3dm 2 года назад

there is a simple solution. just take the log of the expression and you get 1/n (log(1) + log(2) + log(3) + ... + log(n)) - log(n). Using the fact that the integral of log(x) = xlog(x) - x, you can simplify the whole expression to just -1 + 1/n because the logs cancel. So, the log of the expression tends to -1 and therefore the expression tends to 1/e.

@azmath2059 2 года назад

Spectacular limit proof. Thanks for sharing Prof. Penn

@elieabourjeily8520 2 года назад

Using Stirling approximation we have lim n->infinity n!= (2npi)^1/2 (n/e)^n now we get lim n->infinity (n!)^n /n =((2npi)^(1/2)n (n/e)^n/n) / n lim n->infinity (2npi)^(1/2)n =1 (n/e)^n/n = n/e So that lim n->infinity (n!)^n /n= lim n->infinity n/ne = 1/e

@stanleydodds9 2 года назад

Firstly, you haven't accurately stated Stirling's approximation; lim n -> infinity of n! is divergent. What you meant was that n! is assymptotic to the formula given, which can be stated as the limit of the ratio of the two functions being equal to 1. But the main thing here is that Stirling's approximation is a very strong result, and you are using it to prove a much weaker result on the limiting behaviour of n! that doesn't even require the full detail of the assymptotic formula. This is the same as saying "using the fact that the limit is 1/e plus some other information that's even more specific, we have that the limit is 1/e". The proof of Stirling's approximation is harder than that of the limit we are trying to find, so it feels a bit disingenuous to just assume Stirling's approximation is true for this.

@MathSolvingChannel 2 года назад

Stirling's formula + Squeeze theorem 🧐

@AngryArmadillo 2 года назад

He is essentially proving Stirling’s approximation as an intermediate step

@MathSolvingChannel 2 года назад

@@AngryArmadillo right :)

@ymymymymym 2 года назад

What do you think of this straight to the point proof? If Un = (n!)^(1/n)/n ln(Un) = 1/n(ln(1/n) + … + ln(n/n)) with Riemann sum we have lim(ln(Un))=integral from 0 to 1 of ln(x)dx lim(ln(Un)) = -1 lim(Un) = 1/e

@GiornoYoshikage 2 года назад

Absolutely true, it's quick and obvious

@tomctutor 2 года назад

You claim, but have not shown, that the _ℒim_ n→∞: (1/n)(ln(1/n) + … + ln(n/n)) is indeed the stated integral ʃln(x)dx. Michael has pinned this down precisely by the Squeeze Thm. Not saying its not nice, but sometimes simple isn't best.

@ymymymymym 2 года назад

@@tomctutor I think it does not need proof as it is just the geometric calculation of the area under a curve.

@GiornoYoshikage 2 года назад

@@tomctutor the definition of Riemann integral says that it's the limit of Riemann sums when norm of partition goes to 0. It means that the result doesn't depend on partitions we choose, so if function is Riemann-integrable then we can take any sequence of partitions with norm going to 0, and the limit of Riemann sum with the very sequence is equal to the integral itself. That's why we can choose partitions like (0, 1/n, 2/n, ..., n/n) and take the limit which will be equal to the integral. Hope I got the question right

@GiornoYoshikage 2 года назад

@@tomctutor actually my explanation is wrong - ln(x) is not Riemann-integrable on [0;1] because it's unbounded, so there should be another way to prove that solution

@nasrullahhusnan2289 2 года назад

It is simpler to solve in this way: n!=n(n-1)(n-2)(n-3)... =n^n[1-(1/n)][1-(2/n)][1-(3/n)]... The numerator becomes [n^n{1-(1/n)}{1-(2/n}{1-{3/n}...]^(1/n) which simplifies as [1-(1/n)][1-(2/n)][1-(3/n)]... which for n→∞ equals to 1.

@suk-younsuh7322 2 года назад

let A=(n!)^(1/n)/n. ln(A)=(1/n)ln(n!)-ln(n). Note that n-> very large, ln(n!) approaches to n*ln(n)-n. Thus, ln(A)=(1/n){n*ln(n)-n)}-ln(n)=ln(n)-1-ln(n)=-1 Therefore, A=1/e.

@gael8828 2 года назад

Could we have a proof of the stirling approximation for n! when n approches infinity ?

@anarghamondal2036 2 года назад

It is well known actually. Stat Majors have to use that quite frequently I guess. (I am not a Stat Major so I maybe wrong, I am just a high school student).

@dudono1744 2 года назад

stirling approximation is used with Boltzmann distribution because it's about (very) big factorials

@namesurname1982 2 года назад

DuDono, it gives a close approximation since the beginning, just plot a graph.

@user-fi6if8gx3g 2 года назад

Read a wikipedia page about it, there's a proof inside.

@Rose-xe4ct 2 года назад

it’s 2am but im watching a video i dont even understand. this will be useful in the future.

@FF-ms6wq 2 года назад

Great solution Michael.

@CauchyIntegralFormula 2 года назад

If you take the logarithm of the expression, you get (1/n)(sum_{k=1}^n ln(k)) - ln(n) = (1/n) (sum_{k=1}^n [ln(k) - ln(n)] ) = (1/n) (sum_{k=1}^n ln(k/n) ). Then, the logarithm of the limit (which is the limit of the logarithm because log is continuous) is ln(L) = lim_{n->infty} (1/n) (sum_{k=1}^n ln(k/n) ). This limit is a right Riemann sum, so its value is int_0^1 ln(x) dx = -1, and hence the value of the original limit is e^{-1} = 1/e. tl;dr: the expression in the limit is the exponentiation of a Riemann sum, which makes taking the limit equivalent to calculating an integral Fake edit after starting the video: Ah, I see you did something similar, but my manipulation allows the Riemann sum to take place over the same interval every time so you can just wield the "Riemann sums of continuous functions converge" cudgel directly

@Catilu 2 года назад

16:17 blackpenredpen wants to know your location

@ealejandrochavez 2 года назад

There is a much easier solution. Just check that the natural logarithm of the expression we want to compute the limit of is equal to (1/n) [ln(1/n) + ln(2/n) + ln(3/n) + ... + ln(n/n)]. This is a Riemann sum which converges to the integral of ln(x) from x=0 to x=1. The value of this integral is -1, whence the answer is 1/e.

@scratchcriminal 2 года назад

Nice idea but isn’t ln(x) unbounded on [0,1], and thus not Riemann integrable on that interval?

@ealejandrochavez 2 года назад

@@scratchcriminal The Riemann sums can converge to an improper Rimenann integral given reasonable assumptions. For example, finiteness of the Riemann sums, finiteness of the improper integral and monotonicity of the integrand guarantee the convergence.

@lesoleil2465 2 года назад

how i did it : first you prove that lemma : let u(n) be a postive sequence if u(n+1)/u(n) tends to L (a number ) then (u(n))^(1/n) tend to L. then , let u(n)=n! / n^n , and you apply the lemma

@optima4509 2 года назад

Good idea too.

@violintegral 2 года назад

Yep, Michael actually proved and used this exact fact in a previous video to evaluate this same limit. I think it's pretty cool that the ratio and root tests give the same limiting value if the limits exist.

@drorbitaldeathray 2 года назад

A stirling demonstration!

@harshnahata2705 2 года назад

We could use the fact that limit ((n+1)!/n+1^n+1)/(n!/n^n) is 1/e and then take the nth roots which will have the same limit

@bentoomey15 2 года назад

That's a nice trick. I used Stirling's approximation (a big theorem to pull out, but so is the fact the upper and lower Darboux sums bound the integral). Then we have (n!/n^n)^(1/n) = (1/e)*(2pi n)^(1/2n) + O(1/n) Since n^(1/n) ---> 1 and 1/n ----> 0, the limit is 1/e.

@liyi-hua2111 2 года назад

non-log solution from baby Rudin thm 3.37 For any seq of c_n of positive number lim inf c_(n+1)/c_n =< lim inf (c_n)^(1/n) =< lim sup (c_n)^(1/n) =< lim sup c_(n+1)/c_n take c_n = n!/(n^n) then we have lim of ratio parts => sup = inf => limit of root is equal to limit of ratio.

@byronwatkins2565 2 года назад

At 5:00, there is no requirement that x_{n+1} sum ln() looses the relationship between the far left and the far right.

@drissmensouri8624 Год назад

Thanks 👍👍👍

@alexandruvasile9728 2 года назад

This problem can be solved very quickly and easily using the Cauchy-d’Alembert Criterion.

@zakariafifa7048 2 года назад

no you can use stirligue formula log(n!)=n*log(n)-n

@jorgelenny47 2 года назад

I solved it with the fact that lim n->inf of a_n = (a_n+1)/(a_n) by converting the denominator to n^(n*1/n) and grouping the nth root on numerator and denominator. After cancelling most terms you get to lim n->inf (n/(n+1))^n = e^(lim n->inf n(n/(n+1)-1)) = e^-1

@Omar-hm6pu 2 года назад

If we use Reinmann Sums we can just change n! To the product from k=1 to n of k then we change the product to sum introducing Ln and we end up with an improper integral,it’s way easier!

@liyuan-chuanli8468 2 года назад

The radius of the convergence of the series ∑_{n=1}^∞ (n!/n^n) is the limit lim_{n→∞ }(n!)^{1/n}/n by the root test. Then, if you calculate the radius of the convergence of ∑_{n=1}^∞ (n!/n^n) by the ratio test, the computation is easier to of the root test.

@maelhostettler1004 Год назад

Well if you know a bit of theory of asymptotic analysis : n! equiv sqrt(2*pi*n)*(n/e)^n then by special case of conpositon with a power function (works even if dependent of n because its just ln, multiplication and exp) and division of non 0 equivalent : (n!)^(1/n) / n equiv (2*pi*n)^(1/2n)*(1/e) then (2*pi*n)^(1/2n) = e^((1/2n) * ln(2*pi*n)) which by comparison converges to e^0 = 1 so (n!)^(1/n) / n equiv 1/e thus converges to 1/e

@cicik57 2 года назад

yes i also thought that the proof will be through logarifmize and apply the integral definition, here is sort of other solution

@mith_jain_here 2 года назад

Nicee approach... I tried and ultimately got the answer using limit of infinite sum.

@AndrusPr8 2 года назад

I solves this another way. First I went simpler, I did n^(1/n) and solved that the limit to infinity goes towards 1. Then I thought. (n!)^(1/n) is n^(1/n) * (n-1)^(1/n) * (n-2)^(1/n)... etc... each of these individual numbers tend to 1. So the whole (n!)^(1/n) tends to 1. The final factor in the equation is 1/n and this one tends to 0. The whole thing tends to 0. I only calculated 1 limit and reasoned the rest.

@eugeneimbangyorteza 2 года назад

Anyone who knows the Stirling approximation would have a clue on how to answer this

@ssttaann11 2 года назад

The root test for \sum_{n=1}^{\infty} \frac{n!}{n^n}x^n is the limit from the problem. The ratio test for the same series gives 1/e.

@swagatamsen4699 2 года назад

I thought we could do this a bit more simply by 1. Taking log of the sequence first which gives average of terms log(1-k/n) for k=0,1, …, n-1. If the limit of this seq exists and finite then original limit would be exp of it. 2. Since these are left Riemann sum of integral of log(1-x) over 0 to 1, if the integral exists and finite then limit of Riemann sums would be same as the integral. 3. Expanding log(1-x) in power series and integrating by terms we get the series -1/2 - 1/(2.3) - 1/(3.4) - … which is a telescopic sum and yields -1. 4. Thus the original limit is e^(-1)=1/e by conditions of 1. We don’t really need monotone increase of the function, do we?

@dcqin 2 года назад

Nice!

@YorangeJuice 2 года назад

very cool

@serbanpopescu1032 2 года назад

Wow!

@brian554xx 2 года назад

My RU-vid suggestions include what looks like the same limit from a year ago. I assume the result is identical.

@richardheiville937 2 года назад

You assume for your computation that (n+1)^(1/n) has a limit. It's not very rigourous. ln a^b is an ambiguous notation. Anyway it's a nice trick.

@alomirk2812 2 года назад

I just took ln l in bot sides, and beavuse ln is always increasing the ln of the limit equals the limit of the ln, from there you see that as n approachws infinity you get the definition of an integral when f = ln, you solve for the CPV of the int {0,1} ln(x)dx which is 1

@adandap 2 года назад

It's actually -1, but that's the trick I used too. I figured getting rid of the nth root and factorial via a logarithm was going to make life easier.

@comma_thingy 2 года назад

"ln of the limit equals the limit of the ln" is not true since ln is increasing, but rather that it is continuous. Note that floor(x) is increasing, with left discontinuities at integers (which causes the problem) and the limit of 1-1/n is 1, but lim(floor(1-1/n)) = 0 but floor(lim(1-1/n)) = 1

@user-ot4rp8yn8r 2 года назад

When I want to ask this question I dun know how to search, now it just appeared in my recommendation lmao.

@phyarth8082 2 года назад

Stirling factorial approximation formula for factorials.

@snipergranola6359 2 года назад

Using squeeze theorem also helps

@gouharmaquboolnitp 2 года назад

Are you a jee aspirants?

@snipergranola6359 2 года назад

@@gouharmaquboolnitp maths phd

@yakov9ify 2 года назад

But really this is just Stirling's approximation with extra steps so one could just plug it in and use it directly.

@yoav613 2 года назад

I knew this comment will show up,but i think it is very nice to watch this video,because here just with simple tools of integration and algebra and nice trick,we got the answer!

@yakov9ify 2 года назад

@@yoav613 Don't get me wrong I agree that its a nice proof, but its identical to the proof of Stirling's approximation. My point is that if we are gonna do the exact same proof why not just do Stirling's approximation separately and then apply it here.

@yoav613 2 года назад

@@yakov9ify o.k i did not know that,i agree

@MichaelGrantPhD 2 года назад

@@yakov9ify there is far less pedagogical value in just jumping to the answer. Showing the derivation is itself part of the value.

@MichaelGrantPhD 2 года назад

In fact, as someone else pointed out below, he's evaluated this same limit a different way in another video-further reinforcing that it's as much about the technique as the result.

@dudono1744 2 года назад

me: stirling approximation goes brrr

@srittampanigrahi4810 2 года назад

Also we can Do it by using Cesaro stolz theorem.

@scottweisel3640 2 года назад

I don’t know how I can enjoy something so much, while understanding it so little.

@bryanbusby61 2 года назад

Hey Michael, love your work. But, I'm sure you've noticed how there is a black bar on top of the video. Is that a choice you've taken for authenticity? Or something else?

@antoineschuller7527 2 года назад

Question: I found the same limit in 2 min, but I don't know if what I did was rigorous enough to be a 100%-sure result: I took the approximation of n! when n is big (2.pi.n.(n/e)^n) and injected it in the function, getting: n!^(1/n)/n equivalent to (2.pi)^(1/n).n^(1/2n)/e. I checked individually the limits of (2.pi)^(1/n) and n^(1/2n) by using the log and the comparative growth theorem, and I found that both of the logs tend to 0 so both the functions tend to 1, therefore the equivalent tend to 1/e.

@sid025 2 года назад

How do you plot this on graph ?

@SuperSilver316 2 года назад

I think you’ve done this problem before on your channel? But I might be misremembering. You take the natural log of the both sides (assuming the limit exists), and then use L’ Hospital’s Rule once. Once you need to take the limit again the final expression should be digamma(x+1)-ln(x)-1 Using the fact that limit(digamma-ln(x)) = 0, as x goes to inf, and then negate the log with the exponential and you should be good to go.

@violintegral 2 года назад

He did do this limit another way in a previous video, where he first showed that the root test and ratio test both give the same value in the limit if it exists. Since the limit here looks like the root test applied to the sequence a_n=n!/n^n, he then used the ratio test on the same sequence to calculate the limit, which gave the value 1/e. I think both solutions are great and very creative in their own rights. I would love to see some more tough limits on this channel!

@SuperSilver316 2 года назад

Yeah that’s right, I think I posted the same method in the comments just with a little more rigor on my last limit.

@MyNordlys 2 года назад

So the limit of the inverse of the expression = e ;-) Another définition of e ?

@caffreys1979 2 года назад

Exactly my friend 🤠. yes exactly right. lim n >>> infinity of n/((n!)^(1/n)) = e

@meiwinspoi5080 2 года назад

this is an easy limit to calculate with stirlings approximation for n!. since the limit is n tends to infinity stirlings approximation is perfectly legit. the answer can be worked out in 3-4 steps as 1/e quickly. but of course proving stirlings approximation from 1st principles will be an hour long video!!!

@mokshithreddy1634 2 года назад

Love your videos from India 💖❤️💗💛💜💝💞❣️💕

@giuseppemalaguti435 2 года назад

ottimo,stavolta sono arrivato fino alla fine

@alexey_burkov 2 года назад

I'm not quite sure, but probably it could be easier. As long as integral converges, left and right Riemann's sums will tend to actual value of integral. If we logarithm the limit we will get right Riemann's sum, so the natural logarithm of limit equal to integral from 0 to 1 of natural logarithm of x. And that's all. Probably I'm missing something.

@pedroteran5885 2 года назад

It's not Riemann integrable.

@alexey_burkov 2 года назад

@@pedroteran5885 what is not integrable?

@killer408cid 2 года назад

LOL...after plugging in the values 4, 5, 10, and 100, the pattern seemed to be progressing towards 0.35-ish. I literally guessed 1/e. I guess I saved a lot of time. I hope the professor didn't require us to show our work.

@felipelopes3171 2 года назад

L'hopital's rule for the last limit is not sketchy. You prove that the limit of the continuous function is 1, and when that's true, the limit of the sequence given by the values of the function at integer arguments should be the same.

@Bob95051 2 года назад

For those who want to insert Stirling's approximation, I think it's only fair for you to prove it as part of the problem solution. Good luck.

@azmath2059 2 года назад

I agree! Stirlings proof is ridiculously complicated and dificult to follow

@doodelay 2 года назад

So this fact we're using the squeeze theorem for integrals, does it have a name? or is this the actual squeeze theorem itself

@guilhermegomes1314 2 года назад

beautiful

@purim_sakamoto 2 года назад

あ～～　まずそのfactに気付かないとどうしようもないのね　気付いてもむっずかしいなああこれ

@hoangtrile7268 2 года назад

Use stone’ theorem we can prove simpler and general

@doodelay 2 года назад

Why did he choose to integrate ln x in the beginning?

@trevorsesnic8162 2 года назад

Quick question about 13:55. Why is it that when you take the limit of the inequality the less than turn into less than or equal to? I see why it is needed to force the limit to squeeze, but what is the mathematical reasoning? I don't think that this has been discussed in the calc 2 classes that I have taken. Thanks, and it was a great video!

@thoughtfuljanitor6627 2 года назад

You can't actually proove that the inequality stays strict, any proof given of the fact that inequalities are preserved in the limit only proove that non-strict inequalities are preserved in the limit. As an example, consider the sequence (1/n), where n is a positive integer. - It tends towards 0 as n goes to infinity - And yet, 1/n is always strictly greater to 0 In this example, the strict inequality is NOT preserved.

@trevorsesnic8162 2 года назад

@@thoughtfuljanitor6627 I'm a little bit confused by your example. lim n->inf (1/n)=0, yes, and we know that for any number 1/n > 0. I'm a bit confused why this proves that the strict inequality isn't conserved.

@thoughtfuljanitor6627 2 года назад

It's a strict inequality that isn't preserved in the limit. It's a counter-example. Consider the statement: " if a_n and b_n are two sequences such that for all positive integers n, we have a_n < b_n. Then lim a_n < lim b_n" The example I gave prooves this statement is false. If one sets a_n = 0 for all n, then lim a_n = 0, and if b_n = 1/n for all n, then lim b_n = 0. And yet a_n < b_n for all n, so the above statement isn't true (because there are counter-examples). Sometimes, strict inequalities are preserved, but as the above shows, it is not always the case. In the general case, only non-strict inequalities are preserved. (keep in mind a strict inequality can always be generalised to a non-strict: if a_n < b_n then a_n

@syedabuthahirkaz 2 года назад

Dear Mathematician, When A Physicist sees the limit, he immediately knows the limit is 1/e, Thanks to the Stirling's Approximation. But, Mathematician way of unveiling the limit is quite helpful too, to appreciate certain fine points.

@CTJ2619 2 года назад

2 4 6 8 how do exponentiate?

@Icenri 2 года назад

Looks like Stirling formula.

@minwithoutintroduction 2 года назад

عمل رائع . واصل. متتبعك من مدينة تنغير في جنوب شرق المغرب

@fixer3049 2 года назад

I took one look at this and predicted that e would be involved, and the results were as I predicted.

@manucitomx 2 года назад

Something is wrong, I did this by myself. Thank you, professor!

@kevinjohnson4531 2 года назад

Man. Does this thing converge slowly... at n=100 the value is 2.63.

@pawechosta3835 2 года назад

We have got no answer!

@amardexter9966 2 года назад

old video - ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8fI0S-HeYrQ.html

@goodplacetostop2973 2 года назад

Same limit but not the same trick though.

@lexhariepisco2119 2 года назад

i still remember this hahahahahah

@SuperSilver316 2 года назад

I was right!

@atalaki5362 2 года назад

Those are not numbers

@udic01 2 года назад

5:44 its supposed to be ln(n+1) on the last integral. (Edited: i am wrong. Michael was right. My bad)

@violintegral 2 года назад

Nope

@udic01 2 года назад

@@violintegral sorry. My bad.

@violintegral 2 года назад

@@udic01 it's ok lol, it happens

@ieatgarbage8771 2 года назад

Well, n! grows slower than n^n. (n^n)^(1/n) = n (n!)^(1/n) grows slower than n 0?

@cosimodamianotavoletti3513 2 года назад

It is true that n! grows slower than n^n but this isn't enough. In fact by substituting you get 0^0, which is an indeterminate form.

@ieatgarbage8771 2 года назад

@@cosimodamianotavoletti3513 uh, where exactly do you get 0^0?

@cosimodamianotavoletti3513 2 года назад

@@ieatgarbage8771 (n!/n^n)^(1/n) both the base and the exponent tend to 0

@ieatgarbage8771 2 года назад

@@cosimodamianotavoletti3513 ohh I rearranged it to be n!^(1/n)/n, which can be used with l’hop unless n!^(1/n) converges, in which case the answer would be 0 that means the answer is the derivative of n!^(1/n), which I said to be 0. I guess that is just and assumption though. I do know the answer is below 1.

@cosimodamianotavoletti3513 2 года назад

@@ieatgarbage8771 and how exactly would you use l'hop?