thanks viewer for this nice limit!

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Michael Penn

Күн бұрын

Пікірлер: 39

@huyviethungnguyen7788 2 күн бұрын

You can use stolz theorem once again to find the limit at 7:19 if you don't know the stirling's formula.

@slavinojunepri7648 2 күн бұрын

Excellent point

@Happy_Abe 2 күн бұрын

How would you?

@skylardeslypere9909 3 сағат бұрын

@@Happy_Abe what do you mean? Just plug in (n+1) and subtract (n) from it in numerator and denominator

@Happy_Abe 3 сағат бұрын

@@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

@skylardeslypere9909 Сағат бұрын

@@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

@marianaldenhoevel7240 2 күн бұрын

But Professor Penn! How do I know where the good place to stop is today?

@afignisfirer4675 2 күн бұрын

Where is “And This is a Good Place to Stop”???????

@felipevaldes7679 2 күн бұрын

My thoughts exactly

@jbthepianist 14 сағат бұрын

Maybe it wasn’t a good place to stop :(

@kianbla1675 2 күн бұрын

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

@Happy_Abe 2 күн бұрын

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.

@Zaxx70 2 күн бұрын

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.

@rotoboravtov4354 2 күн бұрын

@@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.

@huyviethungnguyen7788 2 күн бұрын

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?

@Happy_Abe 2 күн бұрын

@@huyviethungnguyen7788 oh maybe, I though the limit has to actually exist to a real number

@NullClass 2 күн бұрын

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.

@GiornoYoshikage 2 күн бұрын

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!

@aadfg0 2 күн бұрын

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.

@petersievert6830 Күн бұрын

12:50 So for any alpha > 1/2 we have ß=0 as a solution pair? Seems somewhat plausible anyway...

@Calcprof 2 күн бұрын

I like the re-statement O(numerator) = n^(1/2)

@bot24032 2 күн бұрын

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)

@quite_unknown_1 2 күн бұрын

Exactly my thoughts

@wesleydeng71 2 күн бұрын

I remember that too. Interestingly he did not mention it at all in the video.

@wikipediaboyful Күн бұрын

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!!!!