An arithmetic-geometric limit 

As an interesting consequence, this shows that e^t ≥ t+1 from the AM-GM inequality, with equality if and only if t=0.

There's a solution that's fairly simpler. The GM can be rewritten as (n!)^(t/n). Use Stirling's formula for n!, and because of the t/n exponent almost all factors have a limit of 1, and you're left with an asymptotic equivalent of (n/e)^t for the GM. Next, plug this equivalent into the GM/AM quotient, which is now (n^t / e^t )* (n/(sum of i^t from 1 to n)). Then divide both the numerator and the denominator by n^t, which transforms the AM term into (sum of (i/n)^t from 1 to n)/n, which in the limit is the definition of the Riemann integral of (x^t dx) from 0 to 1. All that's left is to compute the integral, which gives 1/(t+1), and plug that into the quotient, giving (t+1)/e^t.

11:04 Well, "approximating too small" only holds if the integrand is increasing, in other words, if t>0

Speaking of AM and GM, I enjoyed finding the 'hybrid' AGM in the formula for the exact period of a simple pendulum. Would that be worth a look, Michael?

Geomethic Arithmedian.

Shouldn’t our assumptions about t happen when we are looking for the inequalities on the AM. t been lesse than one there flips the inequalities the other way around. I believe what was calculated holds only for t>1

John Cook has done some articles on these topics recently.

I just used Stirling's Approximation up top and approximated the bottom sum by it's integral, which is valid asymptotically. This is simple enough to do in your head.

Can we apply the limit of GM and AM separately (get some kind of infinity) then combine together to get the answer? Like: lim GM = lim n^t/e^t lim AM = lim n^t/(t+1)

I'm never sure if this is serious math, or trolling non mathematicians people.

Fun fact: the natural log of the geometric mean it's the arithmetic mean of the natural logs.

Where is @goodplacetostop ???

22:24 رائع جدا كالعادة

But what about t

this is awesome

Nice video! ❤️

Why not the integral from 0 to n of x^t dt instead of adding the one?

❤interesting 😊

Beautiful

I think I may take up knitting 🙂

What was is used for? Because for a standalone publication is to easy, no?

Amazing🎉🎉🎉So fun❤❤❤