I've never seen anyone write a summation sign with such ferocity

You opened my eyes to new and crazy things!! And that camera angle switch at 6:55 was fresh!

That was cool. I'm happy that I was able to solve this on my own, but only because I was able to leave the answer in function of the sum S = 1 - 1/2 + 1/3 - 1/4 + ..., which I happened to know was ln(2).

Love how simple the final expression is. Btw the audio seems to clip a lot more than in previous videos.

Its crazy that the steps you used were so bizarre and unconventional but it still got you to the same answer! Absolutely brilliant!

Also if you remember the Taylor series of ln(1+x) and vigilant enough, you could easy avoid like last 5 steps.

I noticed very quickly that the integral from 1/n to 1/n+1 was just 1/n-1/n+1 with alternating signs starting with negative. So you get -(1-1/2) + (1/2-1/3) - (1/3-1/4) + ... Getting rid of the paranthesis and grouping terms with common denominator you get: -1 + 2/2 - 2/3 + 2/4 - 2/5. Substracting and adding 1, factoring out -2 you get 1-2AlternateHarmonic. Found ln2 by finding that the Maclaurin series of ln(1+x) fit the alternating harmonic series pretty well.

Love this and the use towards the end of the anti-derivative is icing on a cake!

This is amazingly mind blowing stuff, man! Encore, encore, encore...please.

You make it sound like a fine poetry and that is awesome!

I love the way you make your videos. Keep it up, I learned a lot from you :)

Very cool work! As a matter of fact, the function is well integrable in Peano sense, because the discontinuity of the floor function occurs only in a numerable set of isolated points; then we can correctly replace the improper integral with a series whose terms are definite integrals relevant to adjacent unit intervals [k, k+1), for any positive index k. Then the final evaluation of the series is correctly performed by exploiting the convergence condition of alternate sign series with vanishing terms; a quicker approach to the final evaluation is based on the Taylor-McLaurin series expansion of the function f(x)=ln(1+x) for x=1, i.e. f(1)=ln2, that is immediately recognized as the alternate harmonic series.

"And that finishes this video". Have I got the right Michael Penn?

Great Michael, it is a very nice problem, never seen any problem close to this.

I've never seen these tricks before, thank you!

Very nice integral evaluation. i love watching your videos a lot.

The substitution was key here. It made things very clear to see and is clearly a very good technique for these types of problems. Finding the series by integration was also a neat trick.

That's brilliant one! Especially with geometric series representation.

Very interesting. The function graph would also be welcome. Congratulations on the channel.

I love the deadpan "great", keep it up!

رائع جدا كالعادة. أحب هذه التكاملات الغريبة

I think You are using famous math sum, but you resolve sum in the great wonderful manner. Thank you

Gracias por tan interesante ejercicio.

That was incredible! Very cool integral and algebra/calculus substitutions. I also really liked your camera angle switch at 6:55 so I could see the board. You should keep the camera there! And for some more unsolicited but hopefully still welcome advice: I suggest moving your “please like and subscribe” thing to the end of the video so that your promo thumbnails for other videos don't block your grand reveal! I had to scrub back and forth through the video so I could see the solution on the board. Anyway, cool problem!

I've just seen this on my book this morning and yet here I am watching your video because it's much more interesting, maybe if I start working on summable families you may do a video about it too xD! Thank you a lot !!!

THAT'S SO GREAT!

another nice property of this integral is that if you replace 1/x with x^s you get something that's essentially the alternating Riemann zeta function (though it's evaluated at something like 1/s instead of s, and is multiplied by and has added to it a factor of 1/x^2)

You are brilliant.

amazing!!!

Hey, interesting problem. Thanks. You've got a tiny mistake 2:55. You've forgotten du in your substitution.

awesome problem!

Thanks for giving me new insight...........its brilliant...hve subscribed ur channel...

Increíble resolución.

By minute 7, we had an expression that was obvious to anyone who knew the Taylor series of ln (1+x). But rather than saying that, or deriving the Taylor series, the expanation becomes circuitious over its last 5 minutes.

Dude you are awesome

Hey can you make a video on the 4th isomorphism theorem and composition series. Thanks a ton !

Amazing！！！

hi Penn, I got confused about some of the steps. Since you only have the property of conditional convergence, you are not allowed to change the order of summation to guarantee you reach the same number. What you did in your video, you change the order of the summation.

I was about to try the substitition method, but I backed off as the wikipedia article on integration by substitution gives a theorem, which assumes that the integrand is a continuous function on the interval of integration. I suppose the assumption can be relieved at least to piecewise continuity, since your initial and subsequent methods give the same result.

amazing

Loveeeed❤❤❤❤❤

awesome

I get -1+2ln2 for the case of the ceiling function instead of the floor function. Thank you.

You could just use a little bit of reasoning: Between 1/(n+1) and 1/n the function evaluates at (-1)ⁿ which means that this expression is the sun where 1

Fancy new camera angle 👌

Great.

this is so cool

Got the answer as 1-2ln(2) - let's see if it turns out right! Very strange that this crazy power integral involving floors related to logs!

okay, you can move easier and straightvorward way. Basic idea with whole part or fraction part is to split integral on parts where this prackets give the same value. for example, you can split intevral 0..1 on : I(1/2 ..1) of -1 dx + I(1/3 ..1/2) of 1 dx + I(1/4 ..1/3) of -1 dx +...= -x evaluated in (1, 1/2) +x | (1/2..1/3) - x(1/3, 1/4)... = -1 + 1/2 + 1/2 -1/3-1/3 + 1/4 + 1/4 = the sign change harmonic series is well known and is ln2, so -1 + 2(1-ln2) = 1-2ln2

I forget the problem every time he turns around and I see his back muscles.

lol, I would have done the same thing. except that I would have made at least 3 mistakes with plus and minus. Great video, although your audio is a little much, maybe you can invest into a new micro (a ball like bprp?).

that was cool

10:58 "because this is absolutely convergent" Is it? If you consider the absolute sum you just get 1/n which diverges.

I got stuck at 10:55 when you said "this is absolutely convergent". It looks like the series in question is just a different form of sum(0,inf,(-1)^n/(n+1)), which is an alternating harmonic series, and thus not absolutely convergent.

Great Content. , I should have watched those when I began my math bachelor. Now I am finished soon and I still lack basic stuff in calculus 😁

(i didn't learn really well at math which sadly was my major.) would you mind to tell me: 1. how can "sum((-1)^n*(1/n-1/n+1)),n=1 to inf" split into two parts? from what i recall, alternating series can converge to different result by proper rearrangement. Won't splitting things up make a different result? 2. Considering "integral (-1)^floor(1/x), from 0 to 1" as an alternating series formed by area pieces, integration seems to be no order to converge to a certain answer, or are there any conventions or theorems for this kind of integration? it would be grateful if you could reply to my questions.

mannnnnnnnnnnnnnnnnnnnn this is amazingggg

Great

How does the geometric series converge at the right endpoint of the interval x=1? Shouldn’t |x| < 1?

Nice new integral form. Here is an extension: Do the result (I have not) for [1/x^n] inside the integral. May be interesting.

At the end, you should substitute x=1 first to the expression 1-2lnx, right?

I'm a tad confused by one step. Why is turning floor(u) to a constant valid? Wouldn't u take on the value of n + 1 at a single point in the integration?

Yay! I arrived at the same answer. I wanted to check if I was correct on wolfram alpha, but wolfram got the wrong answer lol

i lost it at 9:21 when that 2 appeared

I would write it as 1-ln(4) but very good video’

Math is crazy!

How do you know that the sequence $\sum_{n=1}^\infty \int_n^{n+1}(-1)^u {1\over u^2}du$ converges to the same number as $\int_1^\infty (-1)^u{1\over 1}du$ does? I mean, the first one is a subsequence from the second. This doesn't mean that if de subsequence converges the main sequence converges too or, in case both converge, share de same limit.

Oh, I hate it so much, when people write a summation sign in a hurry (or from an uncomfortable angle) that it looks like the thing at 6:27. But great video, though!

Can you actually separate the sum if the series doesn't converge absolutely?

But (-1)^(n+1)/(n+1) equals -1 (not 1) when n = 0 as you say @8:51 ...but parenthesis is then not zero later (you mistakenly substract 1 in parenthesis and add 1 outside). Check out the missing "-" on your formula @9:21 (and remove your trick adding and substracting 1).

The correct answer, 1 - 2 ln 2, is negative because the integrand is -1 on the entire interval 1/2

👍👍

I broke up into subintervals without the substitution and got the same sum without having to integrate 1/u², just 1 and -1.

nice problem

feel like I need a cigarette after that.

Didn't you forget the du when doing the u-substitution?

Maple Calculator evaluates this definite integral as 1, without the 2 ln(2) term

Can you explain 10:00 why we did that?

The power n of -1 is either 1 or -1, why their sum is not an integer?

But inserting x^(n + 1) to the series means that for that series to be convergent x should be -1 < x < 1. How to prove that x is between 0 and 1 only?

In Wolframalpha ∫ (-1)^floor(1/x) dx from 0 to 1 I got 0.50... which not equal to 1-2ln(2)

I got a little bit confused with 2

i have done this as(i.e with 1/x)it is without substitution and ended up with the alternating series of 1-1+1-1+1...... which is of course not the correct one .Someone pls correct me .

I think your mic is clipping for some reason. might require a replacement

Really nice talk

I got Ln(4) - 1

7:20 Why can you change the order of sum? The series Σ(-1)^n (1/n) and Σ(-1)^n (1/(n+1)) aren't absolutely convergent.

Missed the usual "that's a good place to stop"

I feel that the substitution was totally unecessary. I arrived at the sum on the right at 7:37 doing essentially exactly the same thing but without the substitution.

Wait hold on. 1 - 2ln(1 + 0) = 1, so plugging in the limits of integration should get you (1 - 2ln(2)) - (1) = -2ln(2) rightt?

For the same integral but the roof function instead of floor, the result is the opposite

What a body 👍 , Sirrrrrr 😉 .

I got immediately lost at 10:00

It's dx=-du/u^2

Czy ty zauważyłeś, że w ostatnich sekundach logo twojego kanału zasłania ostateczny wynik?

Why are you adding x around 10:12?

Reordering a conditionally convergent series? Mh...

If the floor is lava, how do we integrate it?

The answer is approximately -0.338629436111989061883446424291635, which I mean, come on,, obviously! ;)

Noice