i wonder why these educational videos get no views, even a stupid game play video can get higher view( those videos aren't fun or educational at all ) your content deserves more views
Just a smooth remark, my dear friend. It's a matter of notation: as infinity is not a number (except in the dual space), you can't stand a limit equal to infinity but tending to infinity. 😏👍
I mean, imagine the fraction n!/(3^n). What does it equal, who knows, just needs to be greater than 0. Now look at the fraction when you increase n by 1 (n!)/(3^n) * (n+1)(3) So for any n>3, this function will grow larger, making the limit infinity as n gets larger and larger. Edit: watching the video, that was the exact thing you took advantage of
correct me if I am wrong but isn't it obvious that the numerator is clearly greater than the denominator when n --> inf therefore implying that it approaches +infinity? Or am I wrong?
Note ... 6^n/3^n < n!/3^n, where n > = 9. Now lim (6^n)/(3^n), as n goes to infinity = limit 2^n, as n goes to infinity, and this limit is infinite, hence the limit ( n!/3^), as n goes to infinity is also infinity .
To make your proof perfectly clean, you should write that you consider the case n>5 otherwise the part you're deleting doesn't exist. Anyway, you don't need to develop everything. Let's call a(n)=n!/3^n. You have: a(n+1)/a(n)=[(n+1)!/n!].[3^n/3^(n+1)]=(n+1)/3. So a(n) is a rising series. You can also say that a(n+1)=a(n).(n+1)/3 => a(n+1)-a(n)=a(n)/3 and since a(n) is rising, so does a(n+1)-a(n). But by the definition of the limit, a series can't converge if the difference between its consecutive values becomes greater and greater. So the series diverges. And since it's rising, its limit must be +inf. Last method, even quicker: The answer was on your tee-shirt from the beginning.
The factorial function is stronger than any power function a^x ,a>1. It can be shown if we calculate the limit of I(n) = x(n)/x(n+1),were x(n) = n!/a^n, a>1. We have : I(n) = (n!/ a^n )/ (n+1)!/a^n+1 = n!/(n+1)!. a = a/n+1 -->0 when n-->∞. Which makes it easy to see that x(n) is strongly decreasing even for the large values of a. To be honest the product a^n. n! is compared with n^n.
It is not. Only continuous functions are differentiable. The factorial function fails that test because it is only defined for integers (non-negative integers, specifically). You can do a bit of mathemagic to come up with something akin to what you're asking by taking the derivative of the gamma function (a continuous version of the factorial function) and evaluating it at integer values.
I would look at the cases wit n>6: n!/3^n=(1*2*3*3*5*6)/3^6*product(i)/product(3) i=7 to n Let C=(1*2*3*4*5*6)/3^6 n!/3^n=C*product(i)/product(3) i=7 to n > C*product(6)/product(3) i=7 to n = C*product(6/3) i=7 to n = C*(6/3)^n = C*2^n So lim(n!/3^n) > lim(C*2^n) = C*li(2^n) = infinity As you can see, lim(n!/3^n) diverges at least as fast as C*lim(2^n) with some constant C>0. In the video, it was shown, that lim(n!/3^n) diverges faster than lim(1/27*n), but in fact, it diverges not only faster than a linear term, it diverges even faster than an exponential term with base larger than 1.
i just thought of it like 3^n = 3*3*3.. while as n! = 1*2*3*4*5.. So as only 2 products are less than the 3^n product, n! would take over 3^n quite fast
Stirling's approximation states that ln(n!) is approximately equal to n*ln(n)-n Thus, ln(n!/3^n) is approximately n*ln(n)-n-n*ln3=n[ln(n)-1-ln3], which tends to ∞ as n→∞ Therefore n!/3^n→∞ as n→∞
a neater answer would be to probably to prove Stirling's approximation and then write just as you would say but this is also beautifully elegant Edit:Just realized that I didn't write to prove.
Ну вообще этот предел очевиден без просмотра видео. Факториал каждым следующим N домножает на N предыдущий результат, а знаменатель каждое следующее N домножается на 3. очевидно что при бесконечности множитель будет последний как бесконечность деленная на 3, что есть бесконечность. Т.е. предел бесконечный очевидно.
why lots of limit problems need to assume that HS student passed their analysis class?? No one ever take mathematical analysis, but those limits problems force us to use it if we cannot use L'Hospital. Limit problems is the source of why students hate math. Because math teacher doesn't even teach analysis properly, especially in inequalities. Can you imagine how many students in the class that can make the inequality postulate (n!)/3^n >= (1*2*n)/(3*3*3)?? I can bet my 100 bucks it will be less than 10 in regular classes.
This looks too complicated.....n! is n(n-1)(n-2)....*1, therefore for a very large n,, n! will look like n^n. Then n!/3^n = n^n/3^n = (n/3)^n that tends to infinity when n tends to infinity.
Love your chalks and your math, keep up with the good work!
Your way of explaining this was so beautiful :)
Cool method! You are a great teacher.
Great video to teach how to formalize an intuition.
Incredibly helpful, thank you 🙏
Another great video!
Sweet method. I really liked it.
I really like this exercise, thanks
i wonder why these educational videos get no views, even a stupid game play video can get higher view( those videos aren't fun or educational at all ) your content deserves more views
You are doing well.
Your teaching is excellent and I tell you to make videos about formal proof of limits and continuity involving radicals and denominator
I already have those videos. Thanks
BRO UR TOO GOOD AT TEACING MAN. @@PrimeNewtons
you are the man
Just a smooth remark, my dear friend. It's a matter of notation: as infinity is not a number (except in the dual space), you can't stand a limit equal to infinity but tending to infinity. 😏👍
Nice proof , really !
I love you!
Cool limit! For some real fun, replace the 3 with n. 😊
very very nice.
I mean, imagine the fraction n!/(3^n). What does it equal, who knows, just needs to be greater than 0. Now look at the fraction when you increase n by 1
(n!)/(3^n) * (n+1)(3)
So for any n>3, this function will grow larger, making the limit infinity as n gets larger and larger.
Edit: watching the video, that was the exact thing you took advantage of
correct me if I am wrong but isn't it obvious that the numerator is clearly greater than the denominator when n --> inf therefore implying that it approaches +infinity? Or am I wrong?
Note ... 6^n/3^n < n!/3^n, where n > = 9.
Now lim (6^n)/(3^n), as n goes to infinity = limit 2^n, as n goes to infinity, and this limit is infinite, hence the limit ( n!/3^), as n goes to infinity is also infinity .
To make your proof perfectly clean, you should write that you consider the case n>5 otherwise the part you're deleting doesn't exist.
Anyway, you don't need to develop everything.
Let's call a(n)=n!/3^n.
You have: a(n+1)/a(n)=[(n+1)!/n!].[3^n/3^(n+1)]=(n+1)/3.
So a(n) is a rising series.
You can also say that a(n+1)=a(n).(n+1)/3 => a(n+1)-a(n)=a(n)/3 and since a(n) is rising, so does a(n+1)-a(n).
But by the definition of the limit, a series can't converge if the difference between its consecutive values becomes greater and greater. So the series diverges. And since it's rising, its limit must be +inf.
Last method, even quicker:
The answer was on your tee-shirt from the beginning.
Great.
I usually use Stirling’s formula for these kinds if problems but this is better, especially if we have to prove the limit by definition.
Thanks for another great math video. What about the limit of n!/n^n ?
n^n = n×n×n×n×n×...
Whereas
n! = n×(n-1)×(n-2)×(n-3)×...
n^n is clearly bigger
So this limit approaches 0
ratio test is saviour
The factorial function is stronger than any power function a^x ,a>1.
It can be shown if we calculate the limit of I(n) = x(n)/x(n+1),were x(n) = n!/a^n, a>1.
We have : I(n) = (n!/ a^n )/ (n+1)!/a^n+1 = n!/(n+1)!. a = a/n+1 -->0 when n-->∞. Which makes it easy to see that x(n) is strongly decreasing even for the large values of a.
To be honest the product a^n. n! is compared with n^n.
Sir , kindly make a full course on limit and calculas😢😢😢😢
Sorry, I dont understand the equal. If I delete that terms it shouldn't just be major?
Thank you so much
Hell yea!
COOL
1:10 nice man
is the factorial function differentiable
It is not. Only continuous functions are differentiable. The factorial function fails that test because it is only defined for integers (non-negative integers, specifically).
You can do a bit of mathemagic to come up with something akin to what you're asking by taking the derivative of the gamma function (a continuous version of the factorial function) and evaluating it at integer values.
I would look at the cases wit n>6:
n!/3^n=(1*2*3*3*5*6)/3^6*product(i)/product(3) i=7 to n
Let C=(1*2*3*4*5*6)/3^6
n!/3^n=C*product(i)/product(3) i=7 to n > C*product(6)/product(3) i=7 to n =
C*product(6/3) i=7 to n = C*(6/3)^n = C*2^n
So lim(n!/3^n) > lim(C*2^n) = C*li(2^n) = infinity
As you can see, lim(n!/3^n) diverges at least as fast as C*lim(2^n) with some constant C>0. In the video, it was shown, that lim(n!/3^n) diverges faster than lim(1/27*n), but in fact, it diverges not only faster than a linear term, it diverges even faster than an exponential term with base larger than 1.
i just thought of it like
3^n = 3*3*3..
while as
n! = 1*2*3*4*5..
So as only 2 products are less than the 3^n product, n! would take over 3^n quite fast
Stirling's approximation states that ln(n!) is approximately equal to n*ln(n)-n
Thus, ln(n!/3^n) is approximately n*ln(n)-n-n*ln3=n[ln(n)-1-ln3], which tends to ∞ as n→∞
Therefore n!/3^n→∞ as n→∞
a neater answer would be to probably to prove Stirling's approximation and then write just as you would say but this is also beautifully elegant
Edit:Just realized that I didn't write to prove.
What about using the ratio test
Sadly, the topic this student was dealing with was limits. So I couldn't even mention Ratio test.
♾️
Always remember, the only thing bigger than a factorial is another factorial 😅
Tetration: 👁️
I beg to differ: I think, the "ackermann function" is growing much faster than simple factorial ...
en.wikipedia.org/wiki/Ackermann_function
😇
x^x
@@Cubowave Pentation🗿
@@taito404 septation: (idk if that exists)
stirling approximation!
What about the ratio test? Easier
That limit is just 2 * (3/3) * (4/3) * (5/3) * infinite serie of more than one members
Who else is here from O'leary's Infinite Series Sheet?
me lol
This is so obvious. How could this take more than ten seconds? Let fsubn(n) = (n/3) * fsubn-1). This clearly diverges as n gets big.
Ну вообще этот предел очевиден без просмотра видео. Факториал каждым следующим N домножает на N предыдущий результат, а знаменатель каждое следующее N домножается на 3. очевидно что при бесконечности множитель будет последний как бесконечность деленная на 3, что есть бесконечность. Т.е. предел бесконечный очевидно.
I need 3 sec. to find the result
X!=840 how to solve (not graphically)
Use the gamma function. It is an integral, so you’ll need to look it up. Γ(n) = (n-1)! So just solve Γ(n+1) = 840
No solution
esta muito rápido sem explicar
why lots of limit problems need to assume that HS student passed their analysis class?? No one ever take mathematical analysis, but those limits problems force us to use it if we cannot use L'Hospital.
Limit problems is the source of why students hate math. Because math teacher doesn't even teach analysis properly, especially in inequalities. Can you imagine how many students in the class that can make the inequality postulate (n!)/3^n >= (1*2*n)/(3*3*3)?? I can bet my 100 bucks it will be less than 10 in regular classes.
This looks too complicated.....n! is n(n-1)(n-2)....*1, therefore for a very large n,, n! will look like n^n. Then n!/3^n = n^n/3^n = (n/3)^n that tends to infinity when n tends to infinity.
Ну это очень простая задача.
Я её решил за несколько секунд в уме.
SI può ottenere lo stesso risultato più velocemente sostituendo n! con la Formula di Stirling valida per "n grande"
Who else is here from O'leary's Infinite Series Sheet?