10 Series That You Can Do In Your Head (secret weapon: The List)
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- Опубликовано: 8 сен 2024
- In this video, we will use our secret weapon, the list to solve 10 series,
Series of 1/n! 3:36,
Series of 1/n^(2/3),
Series of 1/n^n,
Series of 1/3^n,
Series of 1/ln(n),
series of 1/sqrt(n^2+1),
series of 1/sqrt(n^3),
series of 1/(0.8)^n,
series of ln(n)/n,
series of n!/10^n,
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blackpenredpen
100/(1-x)
Ln(1) = 0 and since its in the denominator, you would be end up in math jail again.
n=2 :D
Look again, n=2 and not 1
Yeah, but he asked why he didn't go below 2 to start his journey into infinity
@@Otomega1 Guys, look at the (ln n)/n series. The guy is right, he deserves the jail
@@rodrigothomaz6329 What's the problem with this series?
I like how you use "everybody" instead of "everything" -- brings the things to life and gives them personality !
Thanks!!
Tomorrow our professor is giving us a blitz test on series calc 2.
I think you have just saved me.
I'm going to teach Calc II for the first time this summer, I'm here to refresh my memory and to find good tips and tricks to give my students.
Very nice!
I am teaching calc 2 again over the summer. How many weeks do you have for your class?
@@blackpenredpen 4 weeks. It's going to be fast!
Lordoftheflies234 what?!!!!!!! 4 weeks?!!!! Mine is 5 and I think it’s crazy enough. May I ask what school?
@@blackpenredpen A cegep (college) in Quebec city. I'd like to have 5 weeks too!
Good luck sir
That secret weapon used during the competitive exam
10/10 would watch again!
Dr Peyam definitely! Final in 10 days!!
The reason why the summation of 1/(0.8)^n as n goes from 1 to infinity is because the common ratio is greater than 1. In order for a infinite geometric series to converge, r
I finished this class and still come back to these videos time and again just because he's so enjoyable to watch. This guy is truly phenomenal, explaining the toughest concepts like they're child's play. One in a million. Props.
That list is sooo helpful!! Im studying independently to prepare for school and my biggest problem was Series. That list helps so much when I use the Comparison Test!!! Thank you soo much!!
I thought you wanted us to compute the sums. Some sums I can do in my head, but two of them (one of which diverges) are values of the zeta function at non-integral arguments, which I don't know.
7:30-7:37 best part
Thanks!!
Great video as always.👍👍👍
The list is quite the weapon you wish to take with you to a battle(exam).
My precious secret: The telescoping series.
1/(0.8)^n diverges because it is a geometric series with a common ratio of 10/8 which is greater than 1.
4:34 you can feel how badly he wanted to say to what special value it converges...
Kepler Lp e
@@andrewchou3277 e - 1
It start at 2 because ln1 = 0
So if we start at n=1 we start with 1/0 which is undifine
Yup1!!
Very nice!
For (F), I would just use the comparison test against the harmonic series, but shifted by one term:
1/√(n²+1) > 1/(n+1), for n ≥ 1
But ∑₁ºº 1/(n+1) = ∑₂ºº 1/n diverges , therefore, (F) diverges.
Fred
Something similar can be done with some limits and improper integrals. Examples:
Limits: lim (x->+ or -infinity) 2x/sqrt(x^2-3) [considering asymptotic behavior of square root], lim (x->+ or -infinity) (x^2+3x+1)/sqrt(4x^2+x^3-2x^2) [same], lim (x->infinity) (1+5x/(2x^2-x+2))^(6x-7) [there's a tricky shortcut for this one]
Improper integrals: int (from 3 to infinity) (x^3+2)/sqrt(x^8+1) [asymptotic behavior of square root or comparison test], int (from 3 to infinity) exp(3x)/(exp(6x)+5exp(3x)+2) [asymptotic behavior of the sum or comparison test]; hints: (i) int (from a to infinity) exp(-px) converges if p>0, diverges if p1, diverges if p
A) e-1
B) diverges
C) n/(n-1)
D) 0.5
E) diverges
F) diverges
G) converges to?
H) diverges
I) diverges
J) diverges (stirling)
Nice video! Can you make a video about the Kempner serieses? I was always fascinated b them how they can converge when the harmonic series and the series over all prime numbers diverges?
Thank you
Best Video about this title so far 🤝🤝♥️♥️♥️
Great video! Always nice to have a cal 2 refresher :)
Thanks.
Why don't you make a video trying to find out the value of some convergent series. Like letter A that is e-1 by the expansion of taylor series
Hey I have a fun calc 1 problem I’d like u to try!
The curve y = ax^2 + bx +c passes though the point (1,2) and is tangent to the line y=x at the origin. Find a, b, c.
Question: For n>1 ln n is >0. In fact (in general) for n in (1, e) we have ln n in (0, 1)....so how can we say for any n ln n > 1? It is not true...but... for n>e we get ln n > 1!!!
thank u sir, for this, this is helpful to me
Thank you sir
Please bro !!!! Do make next video on 100 DIFFERENTIAL EQUATIONS IN ONE TAKE!!! PLEASE🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Ayush Dwivedi i totally agree man 👍👍
yes please
The List proof plz , and why 1/n is the border btwn convergence and divergence. with thanks
The list... So brilliant to summerise convergent and divergent series in a single line :D Also you explained it like a baws.
For those who don't understand why E) diverges:
The serie is a Bertrand's serie, and we have a = 0 < 1 (from n^a) then it diverges.
Thanks Sir it will definitely help me in my jee advanced preparation 💕💕
Tujh se nhi ho payega chod de kuch nhi ho payega tujh se
Ma baap ka paisa mat kharab kar
I like how my professors doesn't taught me this 🙃
Thank you professor, you are very useful
Thank u dear this will help me in teaching children. I going to teach children in 23 may. Thank u
Who are the children?
@@blackpenredpen 6 to 12 students dear
If there was an esport for math you would be a top player. Bless you master pen
Philosopher: bprp, can we get best friend?
Bprp: no, we have best friend at home.
Best friend at home: 100/(1-x)
Even if you only want to know if the series diverges or converges, you can recognise some of them directly and not only say that they converge but show their limit. We know for example that for any real number x we have: exp(x)=sum(n=1,inf.,x^n/n!). So we know that the first series converges to exp(1)=e.
it should be sum n=0 not n=1 for it to equal e so that series actually converges to e-1 not e
The List:
alternatively we could insert the n^1 or n and assume in the next term of inequality as p>1. In this way:
Ln(n)
Silly poem to memorize the List and divergence/convergence of inverses:
Love is power.
Power is a fact!
Less is power itself.
The inverse until this,
Greater power to itself comes.
Just replace "is" or "itself" by the favorite variable n or x.
Got it! I never got structured text in YT before, not even in Chrome, but for some reason I got it in my Android phone, I used WhatsApp as Draft notepad, copy and paste.
1/((.8)^n) doesn't even approach zero since .8 to any power is a decimal and if u divide any number by a decimal it spits out a larger number so it diverges regardless
Natural log of 1 is zero, the reciprocal of zero is undefined, so to make the infinite series meaningful, all the terms should be defined, which is why the index starts at two and not one
Mnemonic:
L for ln
P for n^p
B for b^n
F for factorial
N for n^n
L P B F N
(um......I dont want to think of some dirty words)
: ))))
Bravo :-)
Many of those given serie's nature cam be conclude by just using Cauchy Condensation
Can you provide the link to proofs of this list please?
I haven't done the whole proof yet but here's a small part: ruclips.net/video/ELRSCt66MQQ/видео.html
wouldn't 1/ln n lead to a problem with n = 1, since ln 1 = 0 and therefore would render the equation undefined?
That's why I started with n=2
@@blackpenredpen I figured as much. Thanks for the challenge question.
@@ariel_haymarket yup!! : )))) Thanks for watching too!
love this
#DIV/0! ln(1)
Yay good list i love it :) my maths tacher gave the same trick! Still happy to have you as a complement :)
This was an amazing video!
You have a distance of 20 meters to the equal parts of each section is equal to x so that the first x---->t then the second x ----->1/2t then 1 / 4t … and so on, Calculate the total distance in terms of X, and the time took?
Excuse me, I didn't undertand the F. The aproximation is right, and for larger numbers sum 1/sqrt(n^2+c) aproximate sum 1/sqrt(n^2) = 1/n, when n go to infinity (with c any constant). But sum 1/n^p diverge only when p is bigger or equal to 1. Until this point, we agree.
But 1/sqrt(n^2+c) is ALWAYS lower than 1/n, although slightly. So, what happens?
You can use n^2>=1 and thus n^2+1=1/sqrt(n^2+n^2)=1/sqrt(2n^2)=1/sqrt(2)1/n which diverges because 1/n diverges.
E can't be the sum from 1 to infinity, because ln(1)=0 so you would have 1/0 which doesn't exist.
Yup!
@@blackpenredpen Hey, I just wanted to say that your videos got me really interested in math. I'm even gonna study it next year. Thank you so much.
N equals one gives us zero on the denominator which is undefined
I don't understand at 8:30 (F). Since √(n²+1) > √(n²) it implies that 1/√(n²+1) < 1/√(n²) which means the series is slightly below the critical value i.e. converges no?
I had the same remark. But you are the first to ask ;)
To understand that, take the behaviour at infinity. As n goes very big, it becomes so much bigger than the 1 that you add to it that it begins to not matter whether you add a 1 or not. So eventually "at infinity", n^2 + 1 ≈ n^2, and so 1/sqrt(n^2 + 1) ≈ 1/sqrt(n^2) = 1/n, the sum of which diverges.
But then you could argue "yes but I have to go to a very big n before the 1 starts to not matter !". The key of that answer is in the fact that the harmonic series diverges : the sum of 1/n goes to infinity regardless of the starting n ; whether you start summing at n = 1, n = 10^68 or whatever else, it will never converge and continue increasing to infinity. Then, you could say that you sum starting at some n0 that is big enough that for every n > n0, n^2 + 1 ≈ n^2. This way, the sum of 1/sqrt(n^2 + 1) starting from n = n0 to ∞ will be roughly equal to the sum of the harmonic series starting from n0, which still diverges. Since you would only be adding positive numbers by adding the terms from 0 to n0 that you didn't sum, the complete series diverges. QED
I tried to keep it simple, but if there's something you didn't understand, feel free to ask. Also if you want to look at the proper, rigorous way to do that, it's called "equivalent series". We say that two series (u_n) and (v_n) are equivalent if their ratio tends to some constant value, ie lim u_n/v_n = C some non-zero constant value. The big takeaway is that if sum(u_n) converges and (u_n) and (v_n) are equivalent series, then sum(v_n) converges too. The contraposition says that if (u_n) diverges and (u_n) and (v_n) are equivalent series, then sum(v_n) diverges too. It's pretty hard to prove actually, and uses epsilon-delta methods IIRC
@@matrefeytontias ok understood ty
@@shlomozerbib388 Something to think about: there are infinitely many sequencies a_n such that n< a_n< n^(1+a) for any a>0. For example n*ln(n), n*ln(n)*ln(ln(n)) etc. So if a_n is slightly larger than n, you dont automatically get convergence for the a_n series. The a_n sequence must be larger enough to be equivalent to n^(1+a).
You can use n^2>=1 and thus n^2+1=1/sqrt(n^2+n^2)=1/sqrt(2n^2)=1/sqrt(2)1/n which diverges because 1/n diverges.
If you take n from 1, then the lnn would be 0
1/0 is undefined
incredible, thanks. plz tell me why your microphone is this big(i feel sorry for hands carrying it ). :p
do more algebra things :) pls
how would I solve integral of dx/(x^2+2x-3)^0.5
ln 0 is infinity, ln 1 is zero. 0+ infinite cannot be defined. That's why it starts from 2
I have a problem with equation F. The denominator (n^2+1)^0.5 gets arbitrarily close to n as n goes to infinity, fine. Since the harmonic series is divergent, this must be divergent, right? But wait. That denominator is ALWAYS infinitesimally MORE than n. So what we have is a series that consists of terms that are of the form 1/n^p, where p MUST be greater than one every time. Does that not suggest this series is convergent? In other words, the harmonic series is the hard boundary of divergence, and this equation is on the convergent side of it, even if only infinitesimally. No?
Also, the explanation for I doesn't make sense in light of the explanation for J. In I, ln(n) is much smaller than n^1 as n goes to infinity, so the limit should go to zero and the series converge. It's the same reasoning used in J. What's the difference?
The limit going to 0 isn't enough for convergence.
You can use n^2>=1 and thus n^2+1=1/sqrt(n^2+n^2)=1/sqrt(2n^2)=1/sqrt(2)1/n which diverges because 1/n diverges.
Wait why does H diverge..according to the list, isn't it bigger than others that converge so it also must converge?
1/0.8 = 5/4 and by geometric series. It diverges.
In the description, the second is the series of 1/n^(2/3)
And it's the same in the video
好棒的list!!
Lim n^n=1 as n ->infinity
Aroonima Sahoo
I think you are thinking about the nth root of n
Is it true that you do not use the comparison test here?
I wait for formul for every sume.
ln(1)=0 and it’s in the denominator and we cannot divide by 0
Please make a video on derivative of x^x by first principal because no body has done it by first principal
Please like my comments if you also want it
Well 7:10, n cannot be 1 because ln(1) = 0, so you would get 1/0, which is undefined.
Factorial ❌❌ Factorio✅✅
we start with ln2 because ln1 is equal to zero and you can not devide by zero
Can you please make video tutorial on series convergence and divergence.
I am so confused.
check out my 100 series
1/(b)^n. b
I'm so confused. This makes no sense to me. So because it's less on the list it's convergent?
My teacher says there is no such thing as "The List" 😢 What is its official name or what theorem(s) is it based on?
I keep failing saying The List 😢
Thank you!
can somebody PLEASE tell me what "The List" is???
I want to learn about the list
And diverge and converge
I need a link pls
I love you😍
Little bit of clickbait. The title seems to imply you can also evaluate the convergent series in your head. I was wondering how to do that for C. Of course you can't.
nike! i wonder if they make brown jackets? 🤔
Why brown jacket?
All these summations are pretty non-controversial. My question is, is the following summation convergent or divergent (n goes from 1 to ∞)?
∑ 1/(n² - 10n + 24)
When you look at n²-10n+24, you know that the product of the roots is 24 and that the sum is 10 so n²-10n+24=(n-4)(n-6). Which means that 1/(n²-10n+24) is not defined at all for n=4 or n=6. So the answer is not that the series converges or diverges, it doesn't even exist. We need to be able to calculate every term of a series.
@@italixgaming915 It is certainly true that the n=4 and n=6 terms are undefined, but does that necessarily mean the summation doesn't exist? In fact, by using partial fractions we can rewrite the summation as (1/(x-6) - 1/(x-4))/10. Then, by expanding the terms, we get (1/10)(-1/5 - (-1/3) + (-1/4) - (-1/2) + (-1/3) - (-1/1) + ...) when everything cancels out except for two terms - including the two undefined values - giving us -9/200 for the sum.
@@zanti4132 1/0 is undefined and 1/0-1/0 is still undefined.
@@italixgaming915 It's debatable. To take a similar situation from calculus, is the integral of 1/x from -1 to +1 equal to zero, or is it undefined?
@@zanti4132 Undefined. I suppose that you say that if you integer 1/x on [-1;-1/n] and [1/n,1] you obtain 0 so this is why you pretend that 0 could be the result but if you integer on [-1;-1/n] and [1/n²,1] you obtain log(1/n²)-log(1/n)=log(1/n) and you have an infinite limit when n ---> infinite.
Sorry, you did say n approaches infinity. Forgive this old man.
in-fi-nit-lee :)
that explanation for F) feels incomplete and unsatisfying. The correct way is to derive the inequality 1/n
Bold of you to assume that I am that smart......
P.S you were wrong mate
divergess...
0.16460844=(x^(x)*1.7^(1.7))/((x+1.7)^(x+1.7)) how to solve step by step?
So im guessing this is how our final will be :)?
U r..?
U guess... :)
1/0🚫
I want to get citizenship or residency - how can you help me please
😍😍😍😍
Late, two days late :(
: ((
@@blackpenredpen I love you anyway
The list is not correct
Which part?
Sorry, my maths is no good. I think you should specify n approaches infinity. Or you should say for large n. I was plugging in some figures but this did not work.
I am a 70 yrs old guy learning a lot of maths from you. I did maths in Uni and 98 % percent forgotten.
@@pooi-hoongchan8680
It's okay. No worry about it. I did mention about as n goes to inf right next to "the list".
And wow I am glad to have you here. Hope you enjoy my channel so far! Thank you.
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