Limit of n!/n^n as n goes to infinity, squeeze theorem, calculus 2 tutorial

limit of n!/n^n as n goes to infinity, plus the list, and squeeze theorem
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[@matthewgiallourakis7645]

When we went over this in my calc 1 class, my professor called this the oreo theorem, and then handed out oreos for the entire class. Seems appropriate!

[@blackpenredpen]

Matthew Giallourakis oh wow nice!! I like that idea!

[@jeremybuchanan4759]

*Zero Stuf Oreo Theorem

[@leif1075]

Isnt a simpler proof the fact that the denominator gets bigger mich faster than the numerator..that should suffice..

[@ims6671]

In my country its known as the sandwich theorem, and for the theorem where a sequence is bigger than another sequence that tends to infinity, its called the pizza theorem. Unfourtunately, our professor didnt give us a pizza or a sandwich :(

[@shmuelzehavi4940]

@@leif1075 Maybe it's simpler but is not formal.

[@zaerrr7870]

i'm preparing for my finals completely with the help of your videos and i really really appreciate the help. you're a genius. thank you for your work!

[@rishavchoudhuri8806]

This is also known as Sandwich Theorem.

[@blackpenredpen]

that made me hungry..

[@adude6568]

In my language I learned it as the plier criterion (literal translation)

[@tanishqkushwaha2505]

Did this type of questions came in iit

[@Botisaurus]

I hope you explain the gamma function soon. Remember learning it at university but never really understood it. I strongly believe that you can teach it. Really enjoy your videos :)

[@ozzyfromspace]

Dude, just awesome! I felt like a kid again watching your video.... :) Something about the whole 'math for the sake of math' vibe you had going. Please keep up the interesting videos. -Float Circuit.

[@blackpenredpen]

I will!!! Thank you!!!!!!!!!!!!!!!!!!!

[@Inspirator_AG112]

I would have used a similar approach initially, expanding (n!/nⁿ) as (1 · 2 · 3 · 4 · ... · n) / (n · n · n · n · ... · n), and then observing how (nⁿ) grows faster than (n!), since there are more linear terms in the expansion of (nⁿ) than (n!), when n approaches ∞... The main difference to my approach, however, would be applying one of these following rules for limits of quotients between two functions: · *If Θ(f(x)) < Θ(g(x)), then the limit of f(x)/(g(x) as x approaches ∞ is 0. (This will be the case for (n!/nⁿ).)* · If Θ(f(x)) > Θ(g(x)), then the limit of f(x)/(g(x) as x approaches ∞ of f(x)/(g(x) is ±∞, depending on the signs of each function's limit. · If Θ(f(x)) = Θ(g(x)), then the limit of f(x)/(g(x) as x approaches ∞ of f(x)/(g(x) is the quotient of their 'asymptotic coefficients'. Basically, what this means for the limits of quotients between polynomials as x approaches ∞ is that when applying one of the limit-of-function-quotient rules I listed, *everything minus the leading term of both the numerator and denominator can be omitted, which is incredibly efficient.*

[@yajurphullera9396]

Dude your videos are really awesome and motivating. Thanks man!

[@blackpenredpen]

Yajur Phullera thank you!!!!!

[@ulasaltunn]

I am really appreciated I couldnt find any explanational video like this on youtube nice job bro

[@user-pl7ch9lo1g]

Your videos are very interesting and useful! And, you know, as you mentioned best friend, the fact and the list I started wondering. Are all your videos/lessons a preparation before some gigantic maths problem to solve witch we will need all this knowledge?

[@blackpenredpen]

My videos are usually supplements to what I teach or just math for fun. : )

[@aubertducharmont]

Great video. Loved your way of doing this. What i did was express the factorial with the gamma function and then differentiated it. Then some trivial work was needed, but i arrived at the same result.

[@crosisbh1451]

I have not even taking calculus yet (However I take Calc I next semester), and I find your videos really easy to understand, except for a few calculus concept

[@blackpenredpen]

I am very glad to hear! best of luck and enjoy your classes in the future!

[@blackpenredpen]

I am very glad to hear! best of luck and enjoy your classes in the future!

[@dagajsb6996]

Your explanation is very simple and easy to understand for even me, Japanese 🇯🇵😆

[@ExplosiveBrohoof]

Consider the following generalization: We have lim n-->inf of (n!*k^n)/n^n, where k is a positive real constant. You looked at the case k=1. In fact, for some values of k, the series diverges while for other values, it converges to 0. For what value of k does the series flip from converging to diverging?

[@isws]

i think its diverging when k > e( euler's constant) cause of the factorial approximation but i dont have time rn to solve it😁

[@PanozGTR2]

@@isws Almost! It's actually still divergent when k = e as well. This is because of the sqrt(n) component of Stirling's approximation. If you add in the sqrt(n), and take the limit of (n!*k^n)/(sqrt(n)*n^n) instead then it converges to sqrt(2*pi) instead when k = e, and otherwise behaves the same (aside from the rate of divergence).

[@sUpErTrEkKiE100]

I had that exact problem on my maths assignement this week :)

[@pratyushsharma2922]

That was the doaremon theme.... 👍

[@blackpenredpen]

喜歡小叮噹的請在這按贊!!!

[@sansamman4619]

blackpenredpen how can I message you privately pls?

[@bearlin6136]

現在都叫"多啦A夢" (透漏出年紀了...)

[@sansamman4619]

i dont have twitter

[@blackpenredpen]

Bear Lin 我也是啊... 唉...

[@MrRyanroberson1]

on the top, you end up with: n^n -(sum of k up to n) *n^(n-1) +(sum of [sum of j] * k) *n^(n-2)... already, sum of sums is roughly sum of quadratic, which is cubic, and this is really difficult to work out because the first term is the only thing of degree n, all after are n+1 good thing for the squeeze theorem

[@ccjcjcjcjcjcjjjcjcjcjcjjjjccjj]

Love the doraemon theme piano

[@blackpenredpen]

thank you!!!

[@austinchen6024]

For this problem is it possible to separate the limit into a bunch of products, into lim(1/n)*lim(2/n)*...*lim(n/n), where all the limits are taking n to infinity, and then you could reduce all but lim(n/n) to 0, and lim(n/n) is 1, and then you could say that the product is 0?

[@SlowSabun]

i have a doubt , we can take log on both sides nd then convert it to a integral , like an infinite sum as integral , nd maybe we get it the ans as e??

[@robinsuj]

We called it "the sandwich rule"

[@GrandAdmiralMitthrawnuruodo]

In my highschool mathematics class I learned that I could do it as follows: Both the top as well as the bottom of the fraction go to infinity, but since the bottom grows so much faster the limit must be 0. Would that argumentation be correct?

[@nerd_gameratg6825]

Very thanks from Brazil!!

[@AdamDavis444]

"The list" is very useful for anyone studying Big O Notation.

[@blackpenredpen]

yes!

[@Theraot]

I am pass all that, but I wonder how my teachers would have reacted if I wrote down "By Best Friend Theorem". I did one solve an integral "By Table" and got a bad grade... I think my table as too good for my teacher's taste. The book was McGraw Hill Schaum's Mathematical Handbook of Formulas and Tables by Murray R. Spiegel - not sure of the edition, it was one of those cyan cover.

[@josemanuelalvarezguzman6330]

I'm a student from Mexico, and I finally found help!

[@Nine-2545]

try solve limit n -> infinity of equations ((n^2)!)/(n^2n)

[@JayTemple]

I thought I was so clever because I looked up Stirling's Approximation for n! and divided by n^n.

[@GlorifiedTruth]

Why do you use a Magic Eight Ball as a mic? (I can't be the first to have said this, I know.)

[@evelocz]

You can use limit comparison on this and compare it to (1/n) *1*1*...*1>n!/(n^n) and as n goes to infinity 1/n goes to zero.

[@evelocz]

I should have actually watched the video before commenting lol. You did exactly what I said

[@t_kon]

Squeeze / LCT can work just fine

[@rodrigosuarezcastano732]

That squeeze theorem really fucks me up, great video thought

[@mihaiciorobitca3343]

i have a question for you black pen red pen,if any positive number devided by inf or negative inf is equals to 0 that means any constant positive number devided by 0 is actually equals to + or - inf ?

[@martinzone8153]

It does not equal, but approaches 0 from positive or negative side. In the same way, you cannot divide by 0, but you can approach 0 as divisor from its negative or positive side for the result to approach negative or positive infinity respectively. The same is valid for the infinity. The zero is the center of the Universe, there is nothing at that point. When u reach it, you don't need math any more, but if you change your mind and turn around, you'll never stop walking.

[@JashanTaggar]

Mihai Ciorobitca you would have to test it using close points to come to a hypothesis near that point

[@younessbou6489]

thank you very much from Germany

[@RICOLINO30]

Crack 👏👏👏 saludos desde Argentina 🇦🇷🇦🇷🇦🇷

[@dogol284]

I wrote a full multi-paragraph comment about how I did it and then I started the video and saw he did it the exact same way.

[@adamkangoroo8475]

I've watched this already but I didn't remember the list, it's great. N-factOREO!

[@reazraza]

I gave this same problem to my fnd today and 3 hrs later you uploaded it😂

[@blackpenredpen]

LOL!!! amazing!

[@reazraza]

RIT123 kzbin.info/www/bejne/bmrHZplujLqZZ7s lol he already made one. Same thing but a little different

[@freddietalbot9189]

im in 2nd year uni doing mathematics and this came up as part of a dual limit question. what have i signed up for ??

[@seroujghazarian6343]

Or, you could just prove with the ratio test that the series \sum_{n=1}^{♾}{n!/n^n} converges, and you get that \lim_{n-->♾}{n!/n^n}=0 on a silver platter using the contrapositive of the test for divergence

[@myREALnameISiAM]

You use 1 over n because, when you compare the factorial series versus the the exponential series, the final term of the equation is always 1 vs n.

[@oers9584]

hello, i love your videos. But is not enough just to use that the limit of products is equal to the products of limits, where lim when n goes to infinity from 1/n is zero ?

[@reazraza]

You are really good you should make a video on fibunacci sequence

[@blackpenredpen]

I already did! kzbin.info/www/bejne/d2bXc6muebJsjJY

[@reazraza]

Oh thanks

[@jordilleixalopez7421]

Can you do the limit when x goes to inf of: x!^1/x - (x-1)!^1/(x-1) ?

[@redsalmon9966]

Sad that Doraemon is not a thing in the US, hope that Doraemon will get popular in the US

[@zazkegirotron]

Red Salmon sadly its too old to even have a chace

[@redsalmon9966]

It's old, but classic and wellmade

[@zombiesalad2722]

That show was my childhood

[@redsalmon9966]

Zombie Salad *SAAAAAME!!!!*

[@Teknorg]

Please do a limit n!!/ n^n (Two times factor)!

[@bullinmd]

"The List" appears to correspond to Big O notation. The most desirable Big O in algorithms correspond to the leftmost items on the list.

[@joekerr3638]

Yep, rates of growth

[@nishanthapradeep9003]

this is so helpful thanks bro

[@Gamer45507]

How to do for sequence ???

[@alexandre4393]

Sterling : n! ~ √(2π n) * (n/e)^n ==> n!/n^n ~ (√(2π n) * (n^n/e^n)) / n^n n!/n^n ~ √(2π n)/e^n n!/n^n ~ √((2π n)/e^(2n) ) lim =0 (comparative growth)

[@jesuspinales8738]

What is the limit as n goes to inf of ln(n!)/ln(n^n)?

[@MojeZycieZeMna]

you are so cool, thank you for beeing. because of you I don't have to beg my parents to pay my course retake fee

[@williamnathanael412]

For the list, where does the double exponential fit? I mean, smt like a^b^n

[@physicsphysics1956]

2^2^x>2^x^2=2^(x*x)=(2^x)^x>x^x

[@mallakbasheersyed1859]

At 5:33 he said why he did not take 1/n as 1, can somebody explain,with the same argument 2/n can't be treated as 1 ,explain clearly

[@physicsphysics1956]

do formula for a cubic equation

[@habloia]

Your demostration is amazing, I had to solve it using the D'Alembert criteria the which is not as funny as this metod!

[@kodx9995]

thanks so much !

[@johnfraser8116]

Cool. Thanks!

[@lukasnitsch8641]

flawless proof

[@blackpenredpen]

thank you!!!!

[@82rah]

Is using Stirling's formula for n! (via the Gamma function for n) a legitimate way to do this limit?

[@ianmoseley9910]

82rah Yes, but he specifically addressed that at the start of the video

[@stephenbeck7222]

Yes but he wanted this to be more of a precalc or early calculus limit. Gamma function and Stirling formula are more advanced.

[@Hamidbinsuhail]

Its also called sandwich theoram

[@alexjosephpius5893]

The answer should be 1/e >>> In the second step take log(n!/n^n)

[@yuvalpaz3752]

you can prove this straight from the definition of small-o: we need to show that for all k>0, n>a, n!=1 it is obvious from the definition of n! and n^n if k

[@JustSimplySilly]

Is it possible to calculate the series of the function from n=1 to infinity?

[@t_kon]

JustSimplySilly power series probably? It does converge though

[@verainsardana]

Not all, of some series

[@user-jj8dd5vo6b]

Thanks for video, it's very useful, helped me. And interesting fact. In Ukraine we call theorem that you used "theorem about 2 policeman"

[@Salsh191]

that mister meseeks moment when he finishes the task and disappears 10:40

[@spencerhowell4528]

Amazing video 🙏

[@sergipousmateo2837]

blackpenredpen when you can, check this limit pls. Because i can't understand why: lim x->infinity (2^(2x))!/((2^(2x)-2^x)!*((2^(2x))^(2^x)) = 1/sqrt(e) Thanks for all of your videos, you are awesome!!!!

[@chifengchen9369]

Thank you!

[@blackpenredpen]

You're welcome!

[@lukeLego]

But isn't 1/n also less than or equal to 1?

[@user-qz6zu6ir4e]

Gréât vidéo as always

[@captain_ali_01]

How is 2/n less than or equal to 1?

[@priyaljain5053]

In India, we call it the sandwich theorem

[@nerd_gameratg6825]

Brazil too

[@wkingston1248]

Time to change your name to black pen red pen blue pen.! XD

[@jadegrace1312]

WiSpKing I think it's better without the blue pen because whenever he pulls out the blue pen you know it's a very hard problem

[@ilmaio]

You forgot tetration etc... there is so much more greater progressions than the exponentials...

[@s.saberowski]

please i want lim 2^n/n! to infinity?

[@AMAMohamedAurangaseep]

Bprp bro, I have doubt, How to make (2, 3+4i) in graph?

[@flamingpaper7751]

Can you plug i into the equation?

[@dhruvinkakadia1085]

But you could've got the different answer When you got the relation..: ((n-1)/n)((n-2)/n)((n-3)/n).... ((n-(n-1))/n) You could've written it as (1-(1/n))(1-(2/n))(1-(3/n)).... (1-((n-1)/n)) Then put the limit of n as infinity and we would have got 1×1×1×1...×1=1

[@owdq1nrWaZu5pYI6Z7JU]

1^inf is not always 1

[@Shadow4707]

lim n->inf of 1-((n-1)/n) = 1- (lim n->inf of (n-1)/n) = 1-1 = 0.

[@stephenbeck7222]

The last factor goes to zero, not 1.

[@OonHan]

Easy peasy lemon SQUEEZY

[@MsPataso]

Love your videos 👍👍👍👍

[@blackpenredpen]

thank you!!!!!!

[@JashanTaggar]

I feel like solving this could be done intuitively no? the denominator, n^n obviously grows faster than n! because n! will eventually stop growing? so in an infinite limit scenario, you could say the denominator approaches a larger number than n! by far because it will always keep growing. I understand that showing your work and actually evaluating the limit is the best, legit way, but is this wrong? lmk! :)

[@t_kon]

Taggadude But it's not always true isn't it? Like the sum of harmonic series. By intuitively, one could observe that it gets to 0 as 1/n for n = infinity is very small. However this series is in fact.....divergant.

[@franzluggin398]

This would work if n! indeed had a bound. However, as n increases, you add another, _larger_ factor at the beginning, not a smaller one at the end, so it will always keep growing. In fact, one can show that n! grows faster than 10^n, i.e. lim(n->oo) 10^n/(n!) = 0. EDIT: The correct way of saying what I said in my first sentence would probably be: (n+1)! = (n+1) * n!

[@Harrykesh630]

sir you could have taken log of both sides and then exponentiate the result to find the limit

[@Harrykesh630]

according me after taking log this would convert into the definition of integration of ln(x)

[@estebanzd9434]

Do the limit as n goes to infinity of n factoreo to the n

[@zachansen8293]

9:15 the smallest one is C. This is just basic computer science stuff.

[@shahzadarif9101]

Sir the steps where you reduced every term to less than equal to 1, 1/n is also less than equal to one but it was kept as 1/n, can someone kindly clarify this question, plz

[@stephenbeck7222]

He explained it at that step. For the squeeze theorem to give you a final value of the limit, you have to make the limit less than or equal to a number while simultaneously greater than or equal to the same number, and thus equal to that number. So he saved one factor on the right side that he knows goes to zero while knowing the rest of the factors will be less than or equal to 1, so he can say the entire right side goes to 0.

[@purim_sakamoto]

おおー 大どんでん返しがあるのかと思いきや、当然の結論になった（笑）

[@DanBurgaud]

Sequeeze = Squeeze I thought you are gonna SNEEEZE! HAHAHAHAHA!

[@swapnilmane2107]

Explanation is good, But I liked because of doraemon.

[@douglasespindola5185]

Sometimes, I'm afraid of this guy! Hahaha, just kidding! Greetings from Brazil!

[@kryptplaysgames4482]

what about n^^n

[@dot9498]

Thanx

[@superslash946]

Thank you for this xd

[@rahiabrar8400]

How is 2/n less than one?

[@dyer308]

Was that the theme song from doraemon XD

[@blackpenredpen]

yes!!!!!!!!!!!!!!

[@happy8661]

b^n should be bigger than n!.. e.g. 2^3 is bigger than 3!

[@Stefan-ls3pb]

But if n goes to infinity, n! is always bigger than b^n, no matter what is b. 3628800=10!>2^10=1024 if n goes up, n! grows much faster than b^n. proof lim n->infinity (for all terms, i am too lazy to write them down always) b^n/n!=(b*b*b*b*b*...)/(1*2*3*...*(n-2)*(n-1)*n)=b*b/2*b/3*...*b/(n-2)*b/(n-1)*b/n when n goes to infinity b/n goes to 0 (same for b/(n-2) and b/(n-1) ) so we got =b*b/2*b/3*...*0*0*0=0 so n!>b^n

[@ScholarStream_25]

Meaning n^n/n! Would be infinity than

[@gtax1415]

thanks for the amazing video. x^x=y^y x, y>0 this relationship is union of two functions. these two functions intercept at some point. find this point. (this is my own problem) I hope you make a video about it.

[@nicholaskwok4945]

Innit called sandwich theorem?😂

[@matulawa2320]

daaaaamn what a good video man, I imagined the answer was 0, now I'm glad it was true hahaha

[@SimrahJahan]

Doraemonn introooo❤❤