There is a nice trick to calculate this limit.

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[@christopherallen3353]

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

[@alokdhardubey5030]

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]

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]

16:59

[@Robert_H.]

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.

[@criskity]

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]

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

[@qovro]

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

[@brian554xx]

@@qovro Perfect query.

[@user-lh5hl4sv8z]

@@qovro lmao

[@mr.soundguy968]

So you didn't prefer 1 over e

[@elieabourjeily8520]

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]

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.

[@uy-ge3dm]

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.

[@ymymymymym]

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]

Absolutely true, it's quick and obvious

[@tomctutor]

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]

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

[@GiornoYoshikage]

@@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]

@@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]

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.

[@MathSolvingChannel]

Stirling's formula + Squeeze theorem 🧐

[@AngryArmadillo]

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

[@MathSolvingChannel]

@@AngryArmadillo right :)

[@azmath2059]

Spectacular limit proof. Thanks for sharing Prof. Penn

[@suk-younsuh7322]

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.

[@FF-ms6wq]

Great solution Michael.

[@CauchyIntegralFormula]

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

[@gael8828]

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

[@anarghamondal2036]

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]

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

[@namesurname1982]

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

[@user-fi6if8gx3g]

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

[@ealejandrochavez]

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]

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

[@ealejandrochavez]

@@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.

[@Rose-xe4ct]

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

[@drorbitaldeathray]

A stirling demonstration!

[@alexandruvasile9728]

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

[@zakariafifa7048]

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

[@harshnahata2705]

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

[@Catilu]

16:17 blackpenredpen wants to know your location

[@byronwatkins2565]

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]

Thanks 👍👍👍

[@cicik57]

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

[@eugeneimbangyorteza]

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

[@liyi-hua2111]

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.

[@jorgelenny47]

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

[@liyuan-chuanli8468]

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.

[@snipergranola6359]

Using squeeze theorem also helps

[@gouharmaquboolnitp]

Are you a jee aspirants?

[@snipergranola6359]

@@gouharmaquboolnitp maths phd

[@alomirk2812]

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]

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]

"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

[@mith_jain_here]

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

[@ssttaann11]

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.

[@Omar-hm6pu]

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!

[@lesoleil2465]

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]

Good idea too.

[@violintegral]

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.

[@swagatamsen4699]

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?

[@AndrusPr8]

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.

[@YorangeJuice]

very cool

[@sid025]

How do you plot this on graph ?

[@maelhostettler1004]

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

[@dcqin]

Nice!

[@richardheiville937]

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.

[@phyarth8082]

Stirling factorial approximation formula for factorials.

[@bentoomey15]

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.

[@harris5140]

Wow!

[@brian554xx]

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

[@antoineschuller7527]

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.

[@user-ot4rp8yn8r]

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

[@SuperSilver316]

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]

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]

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

[@doodelay]

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

[@srittampanigrahi4810]

Also we can Do it by using Cesaro stolz theorem.

[@bryanbusby61]

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?

[@MyNordlys]

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

[@caffreys1979]

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

[@yakov9ify]

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

[@yoav613]

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]

​@@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]

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

[@MichaelGrantPhD]

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

[@MichaelGrantPhD]

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.

[@doodelay]

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

[@dudono1744]

me: stirling approximation goes brrr

[@scottweisel3640]

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

[@trevorsesnic8162]

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]

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]

@@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]

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

[@alexey_burkov]

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]

It's not Riemann integrable.

[@alexey_burkov]

@@pedroteran5885 what is not integrable?

[@CTJ2619]

2 4 6 8 how do exponentiate?

[@giuseppemalaguti435]

ottimo,stavolta sono arrivato fino alla fine

[@Bob95051]

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]

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

[@guilhermegomes1314]

beautiful

[@hoangtrile7268]

Use stone’ theorem we can prove simpler and general

[@felipelopes3171]

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.

[@mokshithreddy1634]

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

[@meiwinspoi5080]

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!!!

[@killer408cid]

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.

[@kevinjohnson4531]

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

[@Icenri]

Looks like Stirling formula.

[@purim_sakamoto]

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

[@fixer3049]

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

[@manucitomx]

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

[@minwithoutintroduction]

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

[@pawechosta3835]

We have got no answer!

[@amardexter9966]

old video - kzbin.info/www/bejne/bpesYYZjfcqMqLM

[@goodplacetostop2973]

Same limit but not the same trick though.

[@lexhariepisco2119]

i still remember this hahahahahah

[@SuperSilver316]

I was right!

[@syedabuthahirkaz]

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.

[@atalaki5362]

Those are not numbers

[@udic01]

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

[@violintegral]

Nope

[@udic01]

@@violintegral sorry. My bad.

[@violintegral]

@@udic01 it's ok lol, it happens

[@ieatgarbage8771]

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

[@cosimodamianotavoletti3513]

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]

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

[@cosimodamianotavoletti3513]

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

[@ieatgarbage8771]

@@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]

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

[@davidbrisbane7206]

As an aside, its easy to shoe using the fact that the AGM ≥ Gm, that the fact that n is a positive integer that 0 ≤ lim (as n ⟹ ∞) [ⁿ√n!)/n] ≤ 1/2

[@serbanpopescu1032]

Wow!