Limit of n!/n^n as n goes to infinity, squeeze theorem, calculus 2 tutorial
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- čas přidán 20. 08. 2024
- limit of n!/n^n as n goes to infinity, plus the list, and squeeze theorem
the fact: • THE FACT or • the fact, again (with ...
Check out my 100 calculus 2 problems to help you with your calc 2 final: • 100 calculus 2 problem...
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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!
Matthew Giallourakis oh wow nice!! I like that idea!
*Zero Stuf Oreo Theorem
Isnt a simpler proof the fact that the denominator gets bigger mich faster than the numerator..that should suffice..
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 :(
@@leif1075 Maybe it's simpler but is not formal.
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!
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.
I will!!! Thank you!!!!!!!!!!!!!!!!!!!
This is also known as Sandwich Theorem.
that made me hungry..
In my language I learned it as the plier criterion (literal translation)
Did this type of questions came in iit
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 :)
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.*
That was the doaremon theme.... 👍
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?
i think its diverging when k > e( euler's constant) cause of the factorial approximation but i dont have time rn to solve it😁
@@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).
Love the doraemon theme piano
thank you!!!
"The list" is very useful for anyone studying Big O Notation.
yes!
You are a great teacher with a bubbly personality!
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
I am very glad to hear! best of luck and enjoy your classes in the future!
I am very glad to hear! best of luck and enjoy your classes in the future!
喜歡小叮噹的請在這按贊!!!
blackpenredpen how can I message you privately pls?
現在都叫"多啦A夢" (透漏出年紀了...)
i dont have twitter
Bear Lin 我也是啊... 唉...
Dude your videos are really awesome and motivating. Thanks man!
Yajur Phullera thank you!!!!!
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
We called it "the sandwich rule"
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?
My videos are usually supplements to what I teach or just math for fun. : )
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.
I should have actually watched the video before commenting lol. You did exactly what I said
Squeeze / LCT can work just fine
Sad that Doraemon is not a thing in the US, hope that Doraemon will get popular in the US
Red Salmon sadly its too old to even have a chace
It's old, but classic and wellmade
That show was my childhood
Zombie Salad
*SAAAAAME!!!!*
That squeeze theorem really fucks me up, great video thought
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)
I gave this same problem to my fnd today and 3 hrs later you uploaded it😂
LOL!!! amazing!
RIT123 czcams.com/video/89d5f8WUf1Y/video.html lol he already made one. Same thing but a little different
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
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.
Time to change your name to black pen red pen blue pen.! XD
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
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
I had that exact problem on my maths assignement this week :)
Your explanation is very simple and easy to understand for even me, Japanese 🇯🇵😆
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?
I thought I was so clever because I looked up Stirling's Approximation for n! and divided by n^n.
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.
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
1^inf is not always 1
lim n->inf of 1-((n-1)/n) = 1- (lim n->inf of (n-1)/n) = 1-1 = 0.
The last factor goes to zero, not 1.
"The List" appears to correspond to Big O notation. The most desirable Big O in algorithms correspond to the leftmost items on the list.
Yep, rates of growth
I'm a student from Mexico, and I finally found help!
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.
Easy peasy lemon SQUEEZY
I am really appreciated I couldnt find any explanational video like this on youtube nice job bro
Why do you use a Magic Eight Ball as a mic? (I can't be the first to have said this, I know.)
flawless proof
thank you!!!!
You are really good you should make a video on fibunacci sequence
I already did!
czcams.com/video/A5tBvxDM9V4/video.html
Oh thanks
In India, we call it the sandwich theorem
Brazil too
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?
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 ?
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.
Mihai Ciorobitca you would have to test it using close points to come to a hypothesis near that point
that mister meseeks moment when he finishes the task and disappears 10:40
try solve limit n -> infinity of equations ((n^2)!)/(n^2n)
I've watched this already but I didn't remember the list, it's great. N-factOREO!
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! :)
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.
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!
Crack 👏👏👏 saludos desde Argentina 🇦🇷🇦🇷🇦🇷
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.
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??
Sequeeze = Squeeze
I thought you are gonna SNEEEZE!
HAHAHAHAHA!
9:15 the smallest one is C. This is just basic computer science stuff.
doraemon!
yes yes yes!! hehe
blackpenredpen I am a math enthusiast like you
Explanation is good,
But I liked because of doraemon.
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
Looks up squeeze..I literally did LOL
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!!!!
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
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
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.
Thanks for video, it's very useful, helped me. And interesting fact. In Ukraine we call theorem that you used "theorem about 2 policeman"
Very thanks from Brazil!!
Its also called sandwich theoram
You forgot tetration etc... there is so much more greater progressions than the exponentials...
The answer should be 1/e >>> In the second step take log(n!/n^n)
thank you very much from Germany
Is it possible to calculate the series of the function from n=1 to infinity?
JustSimplySilly power series probably? It does converge though
Not all, of some series
Love your videos 👍👍👍👍
thank you!!!!!!
Where is the black pen red pen YAY?
I still have it, dont worry.
b^n should be bigger than n!.. e.g. 2^3 is bigger than 3!
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
Was that the theme song from doraemon XD
yes!!!!!!!!!!!!!!
sir you could have taken log of both sides and then exponentiate the result to find the limit
according me after taking log this would convert into the definition of integration of ln(x)
Can you do the limit when x goes to inf of: x!^1/x - (x-1)!^1/(x-1) ?
But isn't 1/n also less than or equal to 1?
Sometimes, I'm afraid of this guy! Hahaha, just kidding! Greetings from Brazil!
Gréât vidéo as always
Is using Stirling's formula for n! (via the Gamma function for n) a legitimate way to do this limit?
82rah Yes, but he specifically addressed that at the start of the video
Yes but he wanted this to be more of a precalc or early calculus limit. Gamma function and Stirling formula are more advanced.
Doraemonn introooo❤❤
0:06 Doremon intro music?
daaaaamn what a good video man, I imagined the answer was 0, now I'm glad it was true hahaha
Cool. Thanks!
For the list, where does the double exponential fit?
I mean, smt like a^b^n
2^2^x>2^x^2=2^(x*x)=(2^x)^x>x^x
Please do a limit n!!/ n^n (Two times factor)!
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）
Your demostration is amazing, I had to solve it using the D'Alembert criteria the which is not as funny as this metod!
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 ?
this is so helpful thanks bro
Thank you!
You're welcome!
Did I hear Doraemon? 🤣
Zombie Salad u sure did!!
Amazing video 🙏
thanks so much !
Infinity to infinity is infinity. Infinity factorial is also infinity.
Infinity/infinity is 1.
Duh
(Before anyone says it, it’s a joke)
what happened if it didnt turn out to be sandwich? i got n!/a^n 0
it is for series an= n!/a^n
Do the limit as n goes to infinity of n factoreo to the n
Um, lim(ln(n)) = ℵ₀, lim(nⁱ) = ℵ₀, lim(iⁿ) = ℵ₀, lim(n!) = ℵ₀, but limit(nⁿ) = ℵ₁ because this is the next cardinal infinity, there is no 1:1 mapping possible to the natural numbers in Hibbert’s Hotel for lim(nⁿ). ℵ₀ to the ℵ₀ is ℵ₁ by definition. ℵ₀ + ℵ₀ = ℵ₀, ℵ₀ × ℵ₀ = ℵ₀, but raising it to its own power is a new cardinal infinity with no 1:1 mapping to ℵ₀.
Sequezze
please i want lim 2^n/n! to infinity?
Good
do formula for a cubic equation
Innit called sandwich theorem?😂
What is the limit as n goes to inf of ln(n!)/ln(n^n)?