thanks for the great limit!

[@huguesbornet1211]

Loved the comment on the margin “Fermat could not use” ( for his unknown demonstration so far; unless a piece of paper from back then once pops up)

[@winky32174]

I love when difficult limits simplify so nicely!

[@pietropellegrini7926]

I don't know of you remember but you have already done a video on this problem... It was question B5 in the 2004 Putnam

[@MichaelPennMath]

I realized this after recording, but I decided to post since someone emailed the suggestion in.

[@gp-ht7ug]

Me too i noticed it had already been published but I always like this limit

[@gp-ht7ug]

Me too i noticed it had already been published but I always like this limit

[@krisbrandenberger544]

@ 14:14 The numerator should have x^m, not x^(m+1).

[@Hiltok]

Good spot.

[@neilgerace355]

3:42 😆

[@Xv8M4g3r]

Why does this not end up a L=1 by substituting x=1 into the original formula? x^z for x=1 is just 1 regardless of z. So everything inside of the infinite product just cancels to 1 (no longer depending on n). The product of an infinite number of 1s is just 1 so L=1. I've obviously missed something, but I can't see what?

[@LuqmanEinstein]

Professor,Can you talk about geometry problems? I am so weak in that Field, please:)

[@bjornfeuerbacher5514]

He already did lots of geometry problems in older videos, look around on his channel.

[@LuqmanEinstein]

Thank you my brother

[@Noam_.Menashe]

I don't know why but this seems familiar.

[@nirajmehta4264]

i think he did this problem before, or a similar one.

[@rafaelalcala6939]

Can someone explain better why does the series converge ar 14:38 ? Or a better explanation of whats happening hee?

[@chrayma]

So beautiful !!

[@JRabba1995]

Can someone explain why the infinite product cant be evaluated at x=1 and get a limit of 1? I guess it has to do with the continuity but it is not obvious to me why it is not continuous

[@bjornfeuerbacher5514]

If you evaluate it at x = 1, you get an infinite product 1 times 1 times 1 times ..., i. e. 1 to the power of infinity. And that's not defined.

[@JRabba1995]

@@bjornfeuerbacher5514 the exponentiation is performed inside the product and the exponent would also be 1 after evaluation

[@aniruddhvasishta8334]

To add onto another reply, 1^infty is an indeterminate form that can take on different values depending on how the limit is approached. For a canonical example, note that lim n->infty (1+1/n)^n = e. Doing it the usual way you'd think that the part in parentheses goes to 1 and the exponent goes to infinity and you might think it's 1, but that's not how all limits work.

[@JRabba1995]

@@aniruddhvasishta8334 i would agree except that in this case the exponent does not tend to infty if x=1

[@aniruddhvasishta8334]

@@JRabba1995 I'm not sure I understand. There is no x in my limit?

[@viggo2112]

Bros hairline has an infinite limit

[@LucaHerrtti]

Hi prof, I was reading this article An Elementary Proof of the Strong Law of Large Numbers by N. Etemadi, but had troubles to understand from eq 6 on. I was wondering if you could make a video on it!

[@goodplacetostop2973]

16:42

[@CamEron-nj5qy]

Lowercase Pi in the thumbnail had me confused for a second

[@iithomepatnamanojsir]

Very nice sir

[@markmajkowski9545]

The x~1- ^n to infinity seems to require something more formal since 1^N = 1

[@Hiltok]

There is a difference between limit as n goes to infinity of ( limit as x approaches 1 from below of x )^n = ( 1 )^n = 1 and limit as n goes to infinity of ( limit as x approaches 1 from below of of (x^n) ) = 0. Any instance of x is strictly less than 1 despite the limit being equal to one. As the power goes to infinity, anything less than exact equality of x with 1 sees the result go to zero.

[@markmajkowski9545]

Seems to require more than an assertion - plus then this function would seem to have discontinuity at 1 or Lim approaches 1+ which seems of interest.

[@ПендальфСерый-б3ф]

For any x1- , we get 0.

[@Hiltok]

@@markmajkowski9545 At x=1, the product collapses to just 1. i.e. product for n=0 to infinity of (2/2)^(1^n) = (1^1)^N as N goes to infinity = 1 since the base is now exactly 1. For x approaches 1 from above, the base is strictly >1 and the power (x^n) goes to infinity, so the total product blows up to infinity.

[@Happy_Abe]

Can someone explain @15:40?

[@ibaijurado279]

yeah pls

[@idjles]

@@ibaijurado279he breaks it into A/m + B/(m+1) and ends up With well-known (1-1/2+1/3-1/4…=ln 2) (oscillating harmonic series) with some coefficients, and a change of index to get the 1.

[@IoT_]

​@@ibaijurado279you can use PFE \frac{(-1)^{m + 1}}{m(m + 1)} = \frac{-1}{m} + \frac{-1}{m + 1} And after use the Taylor series of ln(1+x) twice. That's it

[@bjornfeuerbacher5514]

Do you know how partial fraction decomposition works? If yes, that step is easily done; if no, it requires a longer explanation. ;)

[@Happy_Abe]

@@bjornfeuerbacher5514 I do, but it’s a sum not an integral

[@OlympiadProblemsolving]

kzbin.info/www/bejne/e3irn56KbbyiiaM

[@trippia6865]

Right now I'm taking a multivariable calculus course in uni and i have a question about this sort of problems in general. How can I decide on what substitutions, change of variables or any changes on the product/sum/integral/limit that I have in a problem like the ones you solve in the channel in order to proceed? While you're solving the problems the changes are so clear and easy to understand because of the structure of the expression but when I'm faced with different problems I don't always know what I should do. Is there a better way to learn how I can think of these changes by myself? Some examples could be like when you integrate under the sum, the 0'th integral or contour integration. How could i determine which technique is the best for the problem other than when there's a very obvious solution?

[@emperorOfMustard]

I knew that e would pop up just from the statement...

[@farfa2937]

Series being able to be doms and subs implies the existance of switch series

[@whonyx6680]

You could have just used the taylor series of the logarithm, no point in transforming it into an integral.

[@divisix024]

To top it off, the way he did it is basically just deriving the Taylor series of the logarithm

[@unknownstoneageman81]

🔥

[@gabrielcampello7836]

great

[@selimakar7201]

Good job Mr Penn

[@markmajkowski9545]

That’s what he said - but I seems to warrant a more extensive proof than the assertion since x less than 1 - raising to infinite power gives 0. That FEELS like an indeterminate form in need of another step or two as proof. If such out there maybe you know where to look. Thanks

[@Cinnabar_Rhubarb]

There's no indeterminate form since, inside the N limit, x is not changing at all. If you had something like (1 + 1/n)^n, then the base is approaching 1 at the same time as the exponent is growing large, which leads to an indeterminate form. Here, x^n is saying "take a fixed base

[@asifkarim75]

Your video always helps me to understand the physics and mathematics, thanks