Homework: switch the quotient in the logarithm and take minus sign out of it so you get -an²ln(((n+1)/n))/(2n+1) or later -an²ln(1+1/n)/(2n+1) then you use put one n inside the logarithm -anln((1+1/n)^n)/(2n+1) additionaly you divide everything by n so you get -aln(1+1/n)^n)/(2+1/n) in the limit case the experession inside the logarithm is the definition of e so the logarathm becomes 1 and 1/n disappears; so the rest is -a/2
Can't you assume that since the limit when alpha is 1/2 is finite the numerator grows like sqrt(n)? In that case when alpha is bigger than 1/2, the limit must be zero since the denominator would grow quicker.
I thought stolz theorem can be applied as long as bn is strictly monotone above some point and l=liman/bn exists, that includes l being negative or positive infinity?
Cleaner way: express the sum at 3:12 as a Riemann sum for (1-x)ln(x) and compute the integral. Details: n^-2 Σ_k (n+1-k)ln(k) = n^-2 Σ ln(k) + (1/n) Σ (1- k/n)ln(k/n) + (1/n) Σ (1-k/n) ln(n) = A+B+C. We have A = n^-2 ln(n!) ~ ln(n/e)/n -> 0 by Stirling, B -> int_0^1 (1-x)ln(x) dx = -3/4 by Riemann sums, and C = ln(n)/n^2 * n(n+1)/2 = ln(n)/2 * (1+1/n) ~ ln(n)/2, so the whole thing is (1/2 - a)ln(n) - 3/4, meaning a = 1/2 and ln(b) = -3/4.
I accidentally came up with an expression inside the root 5 days ago (in the form `1^n * 2^(n-1) * ... * n^1`) and found the asymptotics of its logarithm yesterday. What a coincidence!
Notice that the sum of ln(k!) from 1 to n equals sum of (n+1-k)ln(k). Then, it is easy to estimate the series of ln(x) and xln(x) using integrals (use sandwiche to formalize). With this I found -3/4 +ln(n)/2 - alpha*ln(n), which is coherent with the video. I prefer to avoid the use of specific approximations such as stirling. The integral idea is also much more flexible.
I was watching this while i was running in a treadmill... That was 3 hours ago! Please!!! Please tell me when to stop! My legs hurt! Please tell me when to stop 😭 It hurts so much! Pleaaaaase!!!!
some time ago the limit of (1^1×2^2×3^3×...×n^n)^(1/n²)/√n was evaluated to be e^(-1/4) on this channel. thr product of it and our limit is equal to the limit of (n!)^((n+1)/n²)/n^(1/2+α), or βe^(-1/4). using the fact that the limit of (n!)^(1/n)/n=1/e (which is fairly eay to prove even if you don't know sterling's formula; take the ln of both sides and turn the left side into the integral of ln x from 0 to 1), we get α=1/2, β=e^(-3/4)
Had been a really popular limit question in our coaching.... apeared 3 times 😅 in our mock papers... tho few had memorised the answer after its two appearances 😂
Where is “And Thats a Good Place to Stop”???????
My thoughts exactly
Maybe it wasn’t a good place to stop :(
We have homework for both the 5th term in that limit near the end and for working out solutions for alpha less than 1/2. It’s not time to stop.
But Professor Penn! How do I know where the good place to stop is today?
You can use stolz theorem once again to find the limit at 7:19 if you don't know the stirling's formula.
Excellent point
How would you?
@@Happy_Abe what do you mean? Just plug in (n+1) and subtract (n) from it in numerator and denominator
@@skylardeslypere9909I see. I guess it’s not that hard, but it looks very long so I didn’t work it out to actually see how easy it might be. Thank you
@@Happy_Abe It will likely be a bit of a tediuos calculation, given that there are more terms now but it'll be quite straight forward
Homework: switch the quotient in the logarithm and take minus sign out of it so you get -an²ln(((n+1)/n))/(2n+1) or later
-an²ln(1+1/n)/(2n+1) then you use put one n inside the logarithm -anln((1+1/n)^n)/(2n+1) additionaly you divide everything by n so you get -aln(1+1/n)^n)/(2+1/n) in the limit case the experession inside the logarithm is the definition of e so the logarathm becomes 1 and 1/n disappears; so the rest is -a/2
We can’t solve the case when the limit is negative infinity since then the application of the Stolz-Cesàro theorem wouldn’t have been allowed.
Can't you assume that since the limit when alpha is 1/2 is finite the numerator grows like sqrt(n)? In that case when alpha is bigger than 1/2, the limit must be zero since the denominator would grow quicker.
@@Zaxx70Yes. If we want to make it formal we can multiply and divide by sqrt(n) and split the limit into a product of two limits.
I thought stolz theorem can be applied as long as bn is strictly monotone above some point and l=liman/bn exists, that includes l being negative or positive infinity?
@@huyviethungnguyen7788 oh maybe, I though the limit has to actually exist to a real number
Cleaner way: express the sum at 3:12 as a Riemann sum for (1-x)ln(x) and compute the integral.
Details: n^-2 Σ_k (n+1-k)ln(k) = n^-2 Σ ln(k) + (1/n) Σ (1- k/n)ln(k/n) + (1/n) Σ (1-k/n) ln(n) = A+B+C. We have A = n^-2 ln(n!) ~ ln(n/e)/n -> 0 by Stirling, B -> int_0^1 (1-x)ln(x) dx = -3/4 by Riemann sums, and C = ln(n)/n^2 * n(n+1)/2 = ln(n)/2 * (1+1/n) ~ ln(n)/2, so the whole thing is (1/2 - a)ln(n) - 3/4, meaning a = 1/2 and ln(b) = -3/4.
I accidentally came up with an expression inside the root 5 days ago (in the form `1^n * 2^(n-1) * ... * n^1`) and found the asymptotics of its logarithm yesterday. What a coincidence!
Notice that the sum of ln(k!) from 1 to n equals sum of (n+1-k)ln(k). Then, it is easy to estimate the series of ln(x) and xln(x) using integrals (use sandwiche to formalize). With this I found -3/4 +ln(n)/2 - alpha*ln(n), which is coherent with the video.
I prefer to avoid the use of specific approximations such as stirling. The integral idea is also much more flexible.
I was watching this while i was running in a treadmill... That was 3 hours ago! Please!!! Please tell me when to stop! My legs hurt! Please tell me when to stop 😭 It hurts so much! Pleaaaaase!!!!
12:50 So for any alpha > 1/2 we have ß=0 as a solution pair? Seems somewhat plausible anyway...
some time ago the limit of (1^1×2^2×3^3×...×n^n)^(1/n²)/√n was evaluated to be e^(-1/4) on this channel. thr product of it and our limit is equal to the limit of (n!)^((n+1)/n²)/n^(1/2+α), or βe^(-1/4). using the fact that the limit of (n!)^(1/n)/n=1/e (which is fairly eay to prove even if you don't know sterling's formula; take the ln of both sides and turn the left side into the integral of ln x from 0 to 1), we get α=1/2, β=e^(-3/4)
Exactly my thoughts
I remember that too. Interestingly he did not mention it at all in the video.
Had been a really popular limit question in our coaching.... apeared 3 times 😅 in our mock papers... tho few had memorised the answer after its two appearances 😂
I like the re-statement O(numerator) = n^(1/2)
Aren’t there any more good places to stop?
11:20 He did that limit correct but weong reasoning, a logarithm does not grow faster than a linear.
Yeah i think he just meant to say the opposite
@redpepper74 I would hope so.
My algorithm is just mocking my fear of mathematics now, isn’t it.
May I ask how did you learn the stops creator theorem and, if you’ve not known it before making this video, how did you end up learning the theorem?
Immediately looking at the thumbnail, I thought, Stotz cezaro TH !
Using the Stirling formula was kinda nasty step to be honest but I understand the video would be too long if not
As a DorFuchs enjoyer (not even from a German-speaking place!), I gotta say...
🎶 Wurzel 2 pi n mal n durch e hoch n 🎶
Even more funny:
When he was rapping monotonously:
... Es gibt unendlich viele Primzahlen ...
That's fantastic limit
NGL, That's a groovy limit 😎
Thx viewer ✌️
Nah
Fantastic
asnwer=111 i dx isit