Crazy limit - This looks familiar!

A calculus competition limit problem with an infinite product! Everyone's favorite number is hiding in here somewhere.
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[@blackpenredpen]

You’re welcome!! You earned the shirt! And thanks for the shout out!

[@blackpenredpen]

I am still thinking this problem. I wonder if we can somehow turn that into a Riemann sum and integral then solve.

[@MuPrimeMath]

My guess us that it's not possible in this case, since you would have to take the ln and then there's nowhere to get a 1/n from.

[@brutalxiao8993]

YES, WE CAN.

[@digxx]

What about \log(L) = \sum_{k=1}^n \log(1+(k/n)/n) ~ \int_0^1\log(1+x dx) = \int_0^1 (x dx + O((dx)^2)) = 1/2, as dx is infinitessimal.

[@joanmartinsuarezloaiza2767]

Yes it is possible, applying Riemann-Stieltjes's integral definition, we come up with the integral from 0 to 1 of ln(1+x)d(x^2)=integral from 0 to 1 of 2x ln(1+x)dx=1/2, after applying exponential we have our solution: e^(1/2)

[@digxx]

@@joanmartinsuarezloaiza2767 Can you elaborate on how precisely you come up with int_0^1 ln(1+x) d(x^2) ?

[@Chrisuan]

Mathematictians when they see an infinite product: E

[@DarthRaven9000]

I really like the style of your explanations: clear, concise, and well presented. You are up there in the global maximum of KZbin math channels. :)

[@MyKillzone2]

In german we call the squeez theorem the police theorem for some reason. It's much more funny to just say police this and you have your answer :D My professor always cracked the joke: Have you seen any Policeman converging? I guess it's funny in lectures not in rl. Nice solution btw keep it up!

[@andreapaps]

Nice use of the squeeze theorem :D. The brute force approach is Ln the whole expression then its the sum of ln(1+i/n^2) for i =1 to n then bring in the power series of ln(1+x)... So you get (1/n^2 + 2/n^2 +... n/n^2) - (1/n^4 + 4/n^4 + 9/n^4 + n^2/n^4) + (1/n^6 +8/n^6 +27/n^6 +... n^3/n^6) +... The first bracket is (n(n+1)/2) (1/n^2) the second bracket is (n(n+1)(2n+1)/6)(1/n^4) the third bracket is (n(n+1)/2)^2 (1/n^6) etc the first bracket simplifies to 1/2 +1/2n the second simplifies to 2/n +3/n^2 +1/n^3 the third 1/4n^2 + 1/2n^3 +1/4n^4 etc we notice that all the terms are devided by some power of n except for the 1/2 from the first bracket thus taking the limit as n tends to infinity we get 1/2. Since Ln of the original tends to 1/2 the original itself tends to e^1/2

[@saulmendoza1652]

There's a nice path... Try ln(limit), then Taylor series for ln(1+x), then sum and you're done...

[@Fematika]

Awesome, love seeing clever ways of evaluating limits.

[@blackpenredpen]

I got a request from a viewer but I could not solve. Fematika and Mu prime, are you interested?

[@Fematika]

blackpenredpen I’m interested!

[@leoyang1.618]

This was just incredible, the explanation was very clear, and the solution was very clever.

[@digxx]

Hey, you can also go forward directly: \log(L)= \sum_{k=1}^n \log(1+k/n^2) and then since x-x^2/2

[@MuPrimeMath]

Nice!

[@VibingMath]

Brilliant solution with brilliant t-shirt

[@chirayu_jain]

Hi mak Vinci today I have made a integral battle 😀

[@VibingMath]

@@chirayu_jain Yes i watched it and liked, keep it up! 😁

[@faustoastudillo2468]

I come from bprp, I was the one who wondered if you were able to lick your elbow :v, Now seriously nice video, better person :)

[@benjamingiribonimonteiro9393]

Very nice and clever solution

[@gabrielfoos9393]

very nicely explained, there is another way to solve this, let x be a real number: x-(x^2/2)

[@3manthing]

Very nice solution. I wouldn't have thought of that.

[@redaabakhti768]

underrated channel

[@mihaipuiu6231]

Mr. prof. Mu Prime Math.I like your proof,maby a little bit overwrite, and like Bleckpenredpen ,you write with a pen with INK finished. Thanks for understanding.

[@MathManMcGreal]

Tasty sandwich!

[@joanmartinsuarezloaiza2767]

Hey mu prime math, I could do it applying the Riemann Stieltjes's integral definition, you can notice the next fact, when we apply logarithm we are going to have \lim_{n \to infty} \sum_{k=1}^{n}ln(1+k/n^2) as you can notice here we have our d(alpha) = d(x^2) due to the grade of the denominator, then we will build our integral, we have the definition of the integral like int_{a}^{b}f(x)d(alpha(x))=lim_{n \to infty}\sum_{k=1}^{n}f(delta(x) +a)(alpha(a+delta(x)k)-alpha(a+delta(x)(k-1)), then we have int_{0}^{1}ln(1+x)d(x^2)=1/2 , then after applying exp we come up with e^(1/2)

[@MuPrimeMath]

Interesting! You would have to enumerate more clearly what your function α would be, and what algebra gets that (α_k - α_{k-1}) term multiplied on the outside without changing the answer, but it seems like that would be doable.

[@mathsmen7209]

Another way to solve

[@saulmendoza1652]

Nice limits by the way... Are you reading Demidovich?

[@MuPrimeMath]

I learned calculus using Stewart!

[@user-eh2ec3rn6w]

Nice solutions .

[@gaurangagarwal3243]

But what if n is odd you can't multiply the middle term by any other

[@MuPrimeMath]

Because lim(a×b) = lim(a) × lim(b), we can take the middle term out of the limit and deal with it separately while still pairing the remaining terms like before. But the limit of the middle term by itself will just be 1, so we don't have to worry about it!

[@Alians0108]

@@MuPrimeMath lim(a×b) = lim(a) × lim(b) are there any restrictions to it? I don't think it works in every case; right?

[@MuPrimeMath]

The primary condition is that all 3 limits in the equation must exist

[@angelmendez-rivera351]

n is tending to infinity, so saying n is odd doesn't really make sense.

[@DigitalNomadDev]

Which university are you student?

[@MuPrimeMath]

I am in high school!

[@DigitalNomadDev]

Wow! İts great you are very skilled.

[@rebakhatun3778]

@@MuPrimeMath which is your country???

[@MuPrimeMath]

USA

[@rebakhatun3778]

@@MuPrimeMath oo

[@profesordanielalvarez3498]

If you take more care of your spelling and some things do not explain them on the air, then your videos will look more professional and easier to understand. Thanks for your time.