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?
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
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.
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. 😏👍
This is wrong. A limit can be equal to Infinity, that's the whole concept of a limit. Now the function itself or the sequence, must be approaching infinity and not be equal to Infinity due to the reasons you stated.
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
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.
You're multiplying an ever increasing large number to the numerator but matching it with just multiplying a 3 to the denominator. So, there is no limit to how big the numerator grows while denominator is just 3 more times of the previous denominator. So the result has to grow into positive infinity with no other choice.
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.
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 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
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.
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.
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.
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.
Your way of explaining this was so beautiful :)
Great video to teach how to formalize an intuition.
Cool method! You are a great teacher.
Love your chalks and your math, keep up with the good work!
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
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?
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
Sweet method. I really liked it.
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 .
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
Incredibly helpful, thank you 🙏
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.
Sorry, I dont understand the equal. If I delete that terms it shouldn't just be major?
Thank you so much
I really like this exercise, thanks
You are doing well.
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. 😏👍
This is wrong. A limit can be equal to Infinity, that's the whole concept of a limit. Now the function itself or the sequence, must be approaching infinity and not be equal to Infinity due to the reasons you stated.
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.
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
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.
you are the man
Sir , kindly make a full course on limit and calculas😢😢😢😢
Cool limit! For some real fun, replace the 3 with n. 😊
Another great video!
very very nice.
You're multiplying an ever increasing large number to the numerator but matching it with just multiplying a 3 to the denominator. So, there is no limit to how big the numerator grows while denominator is just 3 more times of the previous denominator. So the result has to grow into positive infinity with no other choice.
Nice proof , really !
I love you!
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.
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 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
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)
ratio test is saviour
Great.
What about using the ratio test
Sadly, the topic this student was dealing with was limits. So I couldn't even mention Ratio test.
6:49 why is the inequality like so, why can it be equal to
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.
1:10 nice man
What about the ratio test? Easier
Hell yea!
COOL
Who else is here from O'leary's Infinite Series Sheet?
me lol
That limit is just 2 * (3/3) * (4/3) * (5/3) * infinite serie of more than one members
stirling approximation!
Also doesnt diverge not mean infinity
♾️
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.
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
Ну это очень простая задача.
Я её решил за несколько секунд в уме.
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
esta muito rápido sem explicar
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.
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.
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?