man anything from calculus 2 gives me good memories of december 2019 and just how great things were before 2020.

@estebanibarra8082

3 жыл бұрын

I'm taking my calc2 class this year and I cannot be more disappointed u_u f*ck corona

@atesemireltutar6661

3 жыл бұрын

I am currently searching for calculus 2 content and it constantly reminds me still how shitty things are in 2021

@eman-id8rw

Жыл бұрын

Good memories and calculus in the same sentence??

I like the questions that are like "Use the ratio test to-" and then immediately answering "convergent" and moving on to the next question.

I'm glad I looked up how to derive the formula for the Poisson distrubution, as that uses many of the similar tricks in this example which makes it simple to get.

The example problems you chose to work out are always so rich with things to learn from them. I always come away smarter after every video!

@blackpenredpen

7 жыл бұрын

alkankondo89 thank you for your strong comment!

@georgehnatiuk5806

6 жыл бұрын

Hello, You have good instincts. Here is a cute problem: Determine the convergence of SUM: n = 1 to n = infinity of | COS(n) | / n^1/3 numerator = absolute value | Cos(n) | denominator = n raised to 1/3 power Then do without the absolute value on Cos GH

@happypiano4810

3 жыл бұрын

Brrrrram, bum, dadaduhdaduhdadada-dadaluh-dundun Brrrrram, bum, dadaduhdaduhdadada-dadaluh-dundun.

@happypiano4810

3 жыл бұрын

Maybe a little easier: Dundun, dadudadududun.

I helped my friends with this problem a while back and it was really fun. I did it a bit differently but got the same result

@golfchan3613

6 жыл бұрын

Great job! That's the beauty of mathematics: if you have a problem and solve it two different ways, it's likely(or definitely) correct.

I just cant believe the dexterity with this guy... holds two pens in one hand a mic in the other. Others have probably commented this same thing on a number of his videos but I hadn't seen it with my own eyes yet, so I had to say it XD Great content!!

Your each video is a masterpiece .Thanks for such content 😊

Your videos are so great!!!! This problem was tripping me up so bad on my calc 3 homework! Thank you!!!!!!!

you're accent is cute and you're very incisive and to-the-point. Thank you for making content!

love these videos man. even though i'm 17 and only doing basic calculus in school with your videos i'm now able to follow along with the solutions you make and get a peak into the harder stuff i'll do later on 👍

@blackpenredpen

7 жыл бұрын

Natanel Sharabi I am glad to hear! You can also check out my site www.blackpenredpen.com for more resources

@Dafty23

2 жыл бұрын

checking in 5yrs later. how's life?

that algebraic transformation to 1 + a/n was pure finesse.

I love you, you're so excited!

@blackpenredpen

7 жыл бұрын

Thanks!

@dmorgan0628

7 жыл бұрын

Hi and thanks for all of your videos on series, I'll be reviewing them over the weekend, I wish you had more :|. Thanks for all hard work! Would you say that you have enough content for Differential equations at least in the beginning for a student? Thanks!

Evaluating this sum is equivalent to evaluating the integral int_0^infinity xe^x/(e^x-x)^2 dx, which seems to avoid any attacks with methods of contour representation.

omg. tysm. I've spent at least 10 minutes trying to solve this one problem and i didnt even see there was a definition of e hiding in there

Thanks for your help! Made my homework question more understandable!

You are insanely smart. I used to know this stuff. thank you for the math lessons

Another way to do it is to see that for n>1, each term is less than 2/n^2, so this sum is at most 1+2*(zeta(2)-1)=2,29

What is it converging to? What is the number ~1.87985?

5:15 as soon as he flipped the fraction I saw where he was going and my mind was blown... I would've never thought of that

188k people had the same problem and he gave a quick and full answer imagine how much time he is saving us collectively

thank you for your good video. it helps me a lot. your smile makes me happy😉

Wish I was so fluent with maths like you are!

Does the ratio test tells us always 100% about the convergence or the divergence of a series?

factOREO

Neatly done! Next question this begs is, what does it converge to? It's probably nothing intelligible.

@spudhead169

3 жыл бұрын

Total Guess: Sqrt(Pi)

@ffggddss

3 жыл бұрын

@@spudhead169 Summed up to n=4, it's already > 1.8 > √π = 1.77245385... Up to n=10, it's 1.879627... The infinite sum appears to be somewhere around 1.87985+ Fred

@atesemireltutar6661

3 жыл бұрын

@@ffggddss 1.87985386217526 says some website.

@atesemireltutar6661

3 жыл бұрын

@@ffggddss Not everything have to be related to pi or e or golden ratio or some magical crack .p

@ffggddss

3 жыл бұрын

@@atesemireltutar6661 When did I ever say it did?? My reply was to a comment by SpudHead, guessing it to be √π. Fred

I'm really starting to enjoy your videos.

@blackpenredpen

7 жыл бұрын

DICKSON PHISTHUR thank you!!!

Thanks a lot - once more! A delightful revelation (or ten) in each video. Why didn't I have you as a math teacher as a kid? OK, you weren't born then, but surely that's a trivial difficulty for math?!

can we get a proof that the ratio test (and other tests) work?

For the last inequality you could say 21/e>1/3 and because 1/e

Thank you this was amazing I really liked this explanation

I'm here at 2.34 am to watch these, your classes are amazing 👌🏻

What is the solution of summation ( N=1 to infinity) n^2 upon n!

this was the bonus question on my calc quiz

That's the same as lim(n -> inf) !n/n!, so that means that lim(!n/n^n) = 1/e², pretty amazing how combinatorics involve e so much.

nice, thank you helped me a lot !!!

Can you show practical application of this? Or it's just calculus exercise?

Very nice explaination sir👍

Better than my professor. Thank you!

I just had my Calculus 2 Final Exam, and a question about improper integral was asked. Question: Which of the following improper integral converges? A. integral from 1 to inf of ln(x)/[ln(x)+x] dx B. integral from 1 to inf of e^(x)/[e^(x)+e^(-x)] dx C. integral from 1 to inf of (x)^(1/2)*e^(x) dx D. integral from 1 to inf of e^(sin x)/x^2 dx E. None of the above I hope you can make video for this question. Thanks :) p.s.: I answered D

@uxxlabrute

7 жыл бұрын

James Tantono .

@blackpenredpen

7 жыл бұрын

James Tantono D is right. Use the comparison test with e/x^2 will do.

@KalikiDoom

7 жыл бұрын

D converges to (4405072002 π)/4128644111 = 3.351933823297...

@ColoredScreens

6 жыл бұрын

Simply from a logical standpoint, e^(sin x) can never be anything greater than e, while the x^2 gets large very fast. Most likely, you know that the integral of 1/x^2 converges, so this is an even smaller function in all cases than 1/x^2. With a little knowledge of how the given functions behave, you can solve without any difficult math

@epicswirl

4 жыл бұрын

@Omar Valentini They do integrals in calc I in america as well it depends on your college or high school curriculum, not your country.

Thank you!!

Anyone knows what number this sum converges to?

This is a great example to use the ratio test, Isn’t it?

So 1^infinity can also approach 1/e

What is this sum to infinity?

You can tell intuitively that the terms get smaller and smaller (note, doesn't necessarily prove convergence, see the harmonic series!) n^n takes n and multiplies it by n, n-1 times. n! takes n and multiplies it by numbers smaller than n, n-1 times. It's the same amount of multiplications but the terms get smaller. Therefore n^n will be bigger for arbitrarily large values.

@stephenbeck7222

7 жыл бұрын

I think you can go further with the intuition and see that this series as n gets larger and larger compares to a "p series" with larger and larger p values. The harmonic series (with p = 1) diverges but any p value > 1 converges.

thank you so so much!!

I could tell it would converge by the fact that factorial has to grow from lower terms and n^n does not - it has n as the base to its own power, which will become larger quickly, overtaking the numerator as low as 2 for n.

instead of ratio test, is it possible to find (n+1)th term - nth term to test for convergence? (where the difference approaches zero) it feels correct...

@blackpenredpen

7 жыл бұрын

JiaMing Lim unfortunate it wouldn't work Example is series 1/n 1/(n+1) -1/n is approaching 0 but that series div

@stephenbeck7222

7 жыл бұрын

You can use a similar test for an alternating series (where each consecutive switches between positive and negative), but as BPRP pointed out, it does not work for series that are always positive or always negative.

@brunolevilevi5054

7 жыл бұрын

doing the ratio test proves that it converges to 1/e or it just proves that it converges to some other number?

Thank you very much sir❤❤❤❤❤❤

Plz make a video on hilbert space

Thank you so much, you saved my exam

And how is that sum, when n=1 up to infinity?! ...

I used an online graphing calculator at desmos.com/calculator to find out whether it converges. I created a function f(x) = sum of all n!/(n^n) where n=1 to x, then I created a function g(x) = f(x)-f(x-1). I assume that if g(x) approaches zero as x goes to infinity then the function f(infinity) , which is the original series, converges. by 100 it already said 0 then it started saying infinity because the values were too high to store a value and it was basically doing infinity - infinity and this calculator calculates that to be infinity.

makes sense and well explained. thanks!

Converges to what value?

I fking love ur vids, please why you cant be my teacher?:(

what if its the other way around? like n^n/n!

Are there any guidelines about when to apply which of these tests and why?

@sircakey-5902

Жыл бұрын

no unfortunately

The value 1/e is also the value of convergence? or we use it only to "pass the ratio test"?

@peterruf1462

3 жыл бұрын

It's only to pass the test. If you start evaluating the sum by hand you start with 1 for n=1 and that is already more than 1/e

What is the limit??? Plzzx derive it.

very interesting, but i didn't know the ratio test method, do you have a video about it?

@helloitsme7553

7 жыл бұрын

Diego Alonso it's basically based off the fact that geometric series only converge when the ratio is less than one. this 1/e was the limit of the ratio it was approaching, so it's an estimate for the ratio and a tool to see if the series converges or not

@lp4969

7 жыл бұрын

HelloItsMe ok, got it thanks a lot!

amazing!

I misread the title and thought the question is to test if the sum is a rational number. (the sum is finite is too obvious) And I thought, this is interesting. Then I realized I misread the title after fast forwarding the video. I decided to try it myself, using similar strategy which is used to prove e is irrational. Turns out I couldn't make it work. I believe it should be irrational, and likely to be transcendental, I'm wondering how to prove that. Or maybe start from sum of 1/n^n and use similar strategy - but is there a proof for this one? The only discussion I found is 1 xkcd forum post echochamber.me/viewtopic.php?t=66371 . Seems that it's still unknown. But most likely it's transcendental and nobody bothered to prove it.

could you make a video explaining what the sum converges to? Wolfram alpha says 1.879, but I'd love to know why :D

@blackpenredpen

7 жыл бұрын

bruno edwards i actually don't know either.

@jchry3688

7 жыл бұрын

blackpenredpen Well acording to a comment, and he use wolframalpha pro, it's in form of a/b pi, which must have been involved in calculus, since the answer is using pi. He said that a = 52949907 and b = 88489346

@balthazarbeutelwolf9097

5 жыл бұрын

Stirlings approximation for factorial has a pi occurring in it, but if you use it to approximate this sum you get a lower limit of about 1.773, not very close to 1.879.

超級感謝幫助很大🥺

Thank you!

Is it possible to find to what does it converge to?

@fountainovaphilosopher8112

7 жыл бұрын

I bet it's some new irrational number.

But what is the answer

you can use root test note that n!^(1/n)=n/e .. for large number ... root test equal zero ,, then converges..

@bjornfeuerbacher5514

10 ай бұрын

"note that n!^(1/n)=n/e" I don't think you are allowed to take that as a given fact here.

We never studied arithmetic serie in algebra when I was in college...I must had missed out on a lot.

@bjornfeuerbacher5514

10 ай бұрын

That's a topic for calculus, not for algebra.

What is infinite sum? 1/(1-1/e) = e/(e-1)

what does it converge to though?

@swuggerman

7 жыл бұрын

We can't determine what it converges to only that it does converge.

@2timotei

7 жыл бұрын

yes but i think there are methods to proove that it converges lower than some value(i remember i had in highschool some *guy name* method that put bounds on limits but did not find them) so i'm asking if we have some upper bounds or estimates yet

@swuggerman

7 жыл бұрын

Well yes you can put bounds and approximate (depends on the applicable test). I don't know if it is applicable for the ratio test or if you can approximate with the ratio test. I do know that if the integral test applied we could approximate what the series converges to same thing with the alternating series test. However, this isn't an alternating series and I dont know how you would go about integrating with something that has a n! in it.

@2timotei

7 жыл бұрын

www.wolframalpha.com/input/?i=sum&rawformassumption=%7B%22F%22,+%22Sum%22,+%22sumfunction%22%7D+-%3E%22n!%2Fn%5En%22&rawformassumption=%7B%22F%22,+%22Sum%22,+%22sumvariable%22%7D+-%3E%22n%22&rawformassumption=%7B%22F%22,+%22Sum%22,+%22sumlowerlimit%22%7D+-%3E%221%22&rawformassumption=%7B%22F%22,+%22Sum%22,+%22sumupperlimit2%22%7D+-%3E%22infinity%22&rawformassumption=%7B%22C%22,+%22sum%22%7D+-%3E+%7B%22Calculator%22%7D

@swuggerman

7 жыл бұрын

Very interesting. I wish I had the ability to know where that value came from, but that is cool.

isn't your "ratio test" called Rule of D'Alembert?

@edgarhakobyan9782

6 жыл бұрын

Claudio Bear & Co. That's right

@Azadpandit1008

3 жыл бұрын

D alembert ratio test is its full name

smacked my self on the forehead for not noticing euler there ;( got stuck after the hardest part and came here smh thank you!

Using Stirling's approximation would also confirm this.

fuck damn man you are sooo intelligence and you literally own my degrees 100%!!!

try putting a (-1) into the first part and then solve.

I wish you had pointed out that the DEFINITION of e is ( 1 + 1/x) ^ x ... also (1 + x) ^ 1/x. Shows why the answer is based on e.

Sir plss n! -: n^n2 solve sir

the flipping trick was really clever, i tip my fedora sir

Thanks bro i saved my viva

@blackpenredpen

4 жыл бұрын

？

Is it enough to say sum(n!/n^n) < sum(n^(n-2)/n^n) = sum(1/n^2)? It seems non-controversial that n! < n^(n-2), so it looks like a legitimate squeeze to me.

@bjornfeuerbacher5514

10 ай бұрын

"It seems non-controversial that n! That may be non-controversial, but nevertheless, you still would have to prove that. ;)

The series converges even absolutely, doesn't it?

Can't you just make comparison test with sum of (1/n) from n=1 to infinity?

@bjornfeuerbacher5514

10 ай бұрын

And what would be the use of that? The series 1/n diverges.

I'm not even ashamed that I approximated n! by Stirling's Approximation and easily showed it converges. Call me a physicist, but we get shit done! (jk, the last part. I still used Stirling)

Didn't hear here about mentioning "D'Alembert criterion"

Where are you from? Китаец?

(Phi^2) + (e^i(pi)) = Phi

log(387661/26)/log(166)

Why can't I use L'Hospital's rule with n?

@blackpenredpen

7 жыл бұрын

Botond Osváth bc d/dn doesn't make sense since n is just a whole number. For LH rule, we have to d/dx

@botondosvath2331

7 жыл бұрын

Thanks :)

@bjornfeuerbacher5514

10 ай бұрын

Additionally, l'Hospital's rule will give the limit of functions, not of _sums_ of functions. So you could use it to get limits of sequences, but not of series.

can you make a video with an integral test?

@varunsrivastava6421

7 жыл бұрын

Its just taking an indefinite integral, so like any of his integral videos but with a limit at the end.

@aaliaank4478

7 жыл бұрын

you see im only 14 thanks! do you have any university level maths books to recommend?? i would be greatful

@varunsrivastava6421

7 жыл бұрын

Stewart's calculus is a classic. I recommend briefly going through the first few chapters about limits before moving on to derivatives. After you get very familiar with those (seriously get comfortable with all of your derivative rules before moving on) you can start learning about antiderivatives (integrals). At this point, you'd have nearly completely Calculus 1. After, you can start Calc 2 which introduces the idea of the Taylor and Maclaurin series. I won't go into either right now, but you will learn the ratio test, integral test, and more during this time.

@aaliaank4478

7 жыл бұрын

Varun Srivastava i already have covered alot of that. differentation and integration. trig functions expo funxtions, partial fractions, binomials.

@aaliaank4478

7 жыл бұрын

Varun Srivastava Many thanks!

Thank you!sometimes your letters are not clear.

yes

I didnt understood why you need to use the fact, n/(n+1) is always less then one in the positives and approaches 1-, so (n/(n+1))^n in the limit is less then one

@bjornfeuerbacher5514

10 ай бұрын

"n/(n+1) is always less then one in the positives and approaches 1-, so (n/(n+1))^n in the limit is less then one" Huh? Why should that follow?!? There are of examples of sequences a_n which are less than 1 and approach 1, but where the limit of the sequence a_n^n is _not_ less than 1, but equal to 1.

i thought root test works better?

@blackpenredpen

5 жыл бұрын

Hmmm, (n!/n^n)^(1/n) = (n!)^(1/n) /n and you will still end up the limit being 1/e < 1, so still conv. but the limit is kinda hard to show.

@Richard-yw9ew

5 жыл бұрын

@@blackpenredpen oh you are right! last night i was thinking about that all night! thank you so much!

So... it converges to......

@swuggerman

7 жыл бұрын

We don't know what it converges to we just know that it does converge. For lots of series it is very difficult or impossible to determine what the series actually converges to.

@PeterBarnes2

7 жыл бұрын

Desmos says about 1.87985386218. It converges remarkably quickly, as well. www.desmos.com/calculator/itvuutmzmt

@swuggerman

7 жыл бұрын

The 100th term is still remarkably small in comparison to an infinite number of terms. I wouldn't say that is telling you what it converges to. I could be completely wrong with my interpretation though just my thoughts.

@PeterBarnes2

7 жыл бұрын

You can look at the link I put there and change the value of a to see that the decimal expansion does not change in the first 12 digits after the 32nd term. I don't know about you, but for a simple equation like that, I call that a pretty fast convergence. To really argue about this, we might consider looking at slopes of secants of the function x!/x^x to see just how fast it actually does converge.

@willyou2199

7 жыл бұрын

n! ~ n^(n+0.5) e^(-n) sqrt(2pi) n!/n^n = n^0.5 e^(-n) sqrt(2pi) e^(-n) rapidly kills even moderately small n like 10, so yeah it does converge fast.

♡♡♡ I LOVE ♡♡♡

strange that n!/(n^n) is essentially geometric for large n

@SSGranor

7 жыл бұрын

Not that strange if you're familiar with Stirling's approximation, which shows that n! ~ (n/e)^n*\sqrt(2\pi*n). en.wikipedia.org/wiki/Stirling%27s_approximation

Could de answer be justified by the fact that n^n grows faster than n! ? High school student here.

@enesgusinjac4825

4 жыл бұрын

Yes but actually no :P

@bjornfeuerbacher5514

10 ай бұрын

No, that would only prove that the sequence n!/n^n goes to zero. But that is _not_ enough to show that the series converges.

math guys be like: you converge + L + ratio