A Brilliant Limit

Such an irony I just solved this question couple hours ago😂

Can we use the Stirling approximation of factorials for this?

indeed it is brilliant

got a math test tomorrow, hope it will help ( yes I rather look at a video i know nothing about ouvre working )

The part with riemann sum is just genius

This has to be one of the most beautiful math ive ever seen

This topic is taught to 12th grade students in india

Absolutely brilliant

Brilliant…

What an absolutely brilliant solution! This is one of the coolest limits I've ever seen, thank you so much for showing this to the world!

thank you for amazing explanation 😊

maybe a stretch, it looks like it can be solved by taking that root

lim{n->°°}(n!/n^n)^(1/n) = lim{n->°°}(n!^(1/n)/n) = lim{n->°°}1*(2^(1/n))*...*((n-c)^(1/n))*...*(n-1)^(1/n)) all of the constants turn into 1, therefore lim{n->°°}((n-c)^(1/n))*...*(n-1)^(1/n)) ~ lim{n->°°}((n-c)^(c/n)) = lim{n->°°}((n+k)^(-k/n)) = lim{n->°°}(n(1+(k/n))^(n/k)) = e^k*lim{n->°°}(n)^(n/k)) = e^k = (e^-c) = 1/e^c

Why does e keep showjng up everywhere

using my methods i just found out that ((loggamma(n+1) - n * log(n)) * (1 / n)) the bigger n is the closer it is to -1 and exp(-1) is e ^ -1 which is 1 / e
so the ans. is 1 / e

That was the question my high school teacher gave me to solve as the first question of definite integral as a limit of sum

The only thing the title was missing was the exclamatory sign ,btw sir your videos really changed my thought process thank u ❤

And why did you use ln and not a general log?

I may be mistaken, but wouldn’t ln(n!/n^n) be a discrete function if we use the normal definition of factorials and not the gamma function definition? Because then you can’t just plug any value for n in 1/n.(ln(1/n) + ln(2/n) + … ln(n/n))

just use stirling lol...but riemann series works as well

Hai fatto capire benissimo la tua esperienza esilarante. Complimenti per il video!

we can use Stirling's Formula approximation for n!. as the limit tends to infinity n! ~ √(2πn)(n/e)^n

Become very easy with stirling formula lmao 😅

Does this work with any logarithm base or does it have to be the natural log?

🥺wooww💝

You could easily use L'Hopital... (ln(1/n) + ln(2/n)....)/n is the infinity upon infinity form

This is a very beautiful limit 😍

I had a big smile when you made the connection between the series and a Riemann sum.

This beautiful limit is literally like one of the starting questions from the infamous black book of Kota

Multiply power to both numerator and denominator which makes denominator 'n', so limit will become zero

The no nut november equation

does this even work though since factorial requires n to be an integer, so n cant continuously increase as required for an integral

You can solve the integral of ln x from 0 to 1 easily by using the inverse function of ln x. The area is actually equal to the negative of the area under the curve of e^x from -infinity to 0, because they are inverse functions, so you can solve that easily to get -1.

What? But n∈ℕ, what are you taking the area from???

Why
Why would you do this

For the first step, if you took the logarithm of both sides with a base other than e, would the answer still work out to be 1/e? (For example, if you took log base 5 of both sides initially instead of ln)

Να εφαρμοσουμε Stolz υπολογιζεται πιο γρηγορα και ευκολα

Stirling's formula enables us to calculate the limit quickly :)

I got also 1/e

I just know that when the function n/(n!)^(1/n) goes to infinity is e

And the function (n!/n^n)^(1/n) can be simplified into ((n!)^(1/n))/n

Now the function is similar to the other limit but just being he reciprocal

If L was being multiplied by ln, why didn’t he divide the right side by ln instead?

götlükler ezberimde

Why not use the Gamma fumction instead of n! - you never said!

Didn't watch the video, but based on Stirling's approximation I'm gonna say it's equal to e

I used the Stirling formula; n!~sqrt(2*pi*n)*((n/e)^n), and i got e^-1 as well.

I thought the result would be 1 because 1/infinity leads to 0 and any number raised to zero is 1

I didn't understand why the area under the curve is equal to minus one.

Good