
English: 
- [Voiceover] We already have a lot
of experience with the geometric series.
For example, if I have the
infinite geometric series
starting at N equals K to
infinity of R to the N,
which would be R to the K
plus R to the K plus one
plus R to the K plus two and keep
going on and on and on forever.
There are a few things we've
already thought about here.
We know what the common ratio is.
So the common ratio,
which is the ratio between
consecutive terms, is going
to be R to the K plus one.
Actually, let me just write
this as R to the N plus one.
I don't want to fixate on
this K right over here.
So R to the N plus one over R to the N.
Well, this is just going to be equal to R.
Anything to the N plus one over
that same thing with the N,
that's just going to be equal
to R, or R to the first power,
and you see that here.
When you go from one term to another

Korean: 
우리는 이미 기하급수에 대해
많이 다루어 보았습니다
우리는 이미 기하급수에 대해
많이 다루어 보았습니다
예를 들어 무한 기하급수 ∑r^n을 보면
예를 들어 무한 기하급수 ∑r^n을 보면
이는 r^k+r^(k+1)+r^(k+2)...이 될 것입니다
이는 r^k+r^(k+1)+r^(k+2)...이 될 것입니다
이는 r^k+r^(k+1)+r^(k+2)...이 될 것입니다
여기에는 이미 알고 있는 것들이 있습니다
우리는 공비를 압니다
연속하는 두 항들 사이의 비율인 공비는 r^(k+1)
연속하는 두 항들 사이의 비율인 공비는 r^(k+1)
r^(n+1)로 쓰겠습니다
여기에 있는 k를 쓰고 싶지 않습니다
그래서 {r^(n+1)}/{r^n}이 되고
이것은 그냥 r이 됩니다
어떤 것의 n+1제곱을 n제곱으로 나누면
그냥 r, 또는 r^1이 나오는 것을 이렇게 볼 수 있습니다
그냥 r, 또는 r^1이 나오는 것을 이렇게 볼 수 있습니다
한 항에서 다른 항으로 갈때

Bulgarian: 
Вече имаме доста опит със сумите
на геометричните прогресии.
Например, ако имаме безкрайната сума от членовете на геометрична прогресия
r^n за n = k до безкрайност ,
което представлява r^k + r^(k + 1)
+ r^(k + 2)
и продължаваме така
до безкрайност.
Вече знаем няколко неща тук.
Знаем колко е частното.
Частното е отношението
между
два съседни члена, то ще бъде
r^(k + 1)...
Ще го запиша като r^(n + 1)...
Не искам да го фиксирам
като k тук.
r^(n + 1)/ r^n.
Това е равно на r.
Всеки израз на степен (n + 1)
върху същия израз на степен n,
това просто е равно на r,
или на r на първа степен,
както можеш да видиш ето тук.
Когато отиваме от един член до друг,

Portuguese: 
Já temos muita
experiência com a série geométrica.
Por exemplo, se tomarmos a
série geométrica infinita
de n igual a k até o 
infinito, de r elevado a n,
que seria r elevado a k
mais r elevado a k mais um
mais r elevado a k mais dois
e assim por diante.
Há algumas coisas que já conhecemos.
Sabemos qual é a razão.
A razão, que é a relação entre
termos consecutivos, será 
r elevado a k mais um.
Na verdade, deixe-me escrever
isto como r elevado a n mais um.
Eu não quero me fixar neste k aqui.
Então r elevado a n mais um 
sobre r elevado a n
será igual a r.
Algo elevado a n mais um, sobre
a mesma coisa elevado a n
sempre será igual a r, 
na primeira potência,
e que você vê aqui.
Quando for de um
termo a outro

Thai: 
เราได้ฝึก
เรื่องอนุกรมเรขาคณิตมากมาแล้ว
ตัวอย่างเช่น ถ้าผมมอนุกรมเรขาคณิตอนันต์
เริ่มจาก n เท่ากับ k ถึงอนันต์ของ r กำลัง n
ซึ่งก็คือ r กำลัง k บวก r กำลัง k บวก 1
บวก r กำลัง k บวก 2 ไปเรื่อยๆ
ตลอดไป
มันมีสิ่งที่เราคิดได้หลายอย่างตรงนี้
เรารู้ว่าอัตราส่วนร่วมคืออะไร
อัตราส่วนร่วม ซึ่งก็คืออัตราส่วน
ระหว่างเทอมติดกัน จะเท่ากับ r กำลัง k บวก 1
ที่จริง ขอผมเขียนอันนี้เป็น r กำลัง n บวก 1 นะ
ผมไม่อยากใช้ k ซ้ำตรงนี้
r กำลัง n บวก 1 ส่วน r กำลัง n
มันจะเท่ากับ r
อะไรก็ตามยกกำลัง n บวก 1 
ส่วนค่าเดียวกันกำลัง n
มันจะเท่ากับ r หรือ r กำลัง 1
และคุณเห็นมันตรงนี้
เมื่อคุณไปจากเทอมหนึ่งถึงอีกเทอม

Thai: 
คุณก็แค่คูณ คุณแค่คูณด้วย r
และนี่คือการทบทวน
ถ้ามันไม่ใช่การทบทวนสำหรับคุณล่ะก็ ลอง
ดูวิดีโอเรื่องอนุกรมเรขาคณิตดู
และสิ่งที่น่าสนใจคือว่า เราได้พิสูจน์
ด้วยตัวเอง ในวิดีโอเรื่องอนุกรมเรขาคณิต
ว่าถ้าอัตราส่วนร่วม ถ้าค่าสัมบูรณ์
ของตัวส่วนร่วมน้อยกว่า 1
แล้วอนุกรมจะลู่เข้า
และถ้าค่าสัมบูรณ์ของ r
มากกว่าเท่ากับ 1
อนุกรมจะลู่ออก
ซึ่งสมเหตุสมผล เราได้พิสูจน์ไปแล้ว
และมันยังตรงกับเหตุผลที่ว่า ดูนะ
ถ้าค่าสัมบูรณ์ของ r น้อยกว่า 1
แต่ละเทอมจะลดลงไปด้วยสัดส่วนเดียวกัน
มันจะคูณด้วยอัตราส่วนร่วมนั้น
และมันจะลดลงไปเรื่อยๆๆ
กระทั่งมันได้ว่า ถึงแม้ว่าตัวนี้จะ
เป็นผลบวกอนันต์ มันจะลู่เข้าหาค่าจำกัด
ทีนี้ พักเรื่องการทบทวนไว้

Korean: 
단지 r을 곱해줄 뿐입니다
여기까지가 복습입니다
만약 이것을 처음 배운다면
기하급수에 대한 동영상을 보고 오는 것을 추천합니다
그리고 여기에서 흥미로운 것은
우리가 기하급수 관련 동영상들에서
공비의 절댓값이
1보다 작으면
기하급수가 수렴한다는 것을 증명했습니다
만약 r의 절댓값이
1 이상이면
기하급수가 발산합니다
이것은 타당하고 증명을 했었습니다
그리고 여기서도 보면
만약 r의 절댓값이 1보다 작으면
각 항이 그 공비만큼 작아질 것이고
다른 말로 그 공비만큼 곱해질 것이고
그 값이 계속계속 작아져서
결국, 이 급수가 무한의 덧셈임에도 불구하고
유한한 값으로 수렴할 것입니다
이제 복습을 끝냈으니까

English: 
you are just multiplying, you
are just multiplying by R,
and this is all review.
If it is not review, I encourage you
to watch the videos on geometric series.
And what's interesting
about this is we've proven
to ourselves, in the videos
about geometric series,
that if the common ratio,
if the absolute value
of the common ratio is less than one,
then the series converges.
And if the absolute value of R
is greater than or equal to one,
then the series diverges.
And that makes sense,
we've proven it as well,
but it also makes
logical sense that, look,
if that absolute value
of R is less than one
then each term here is going
to go down by that common,
or it's going to be multiplied
by that common ratio,
and it's going to decrease
on and on and on and on
until it makes sense
that, even though this
is an infinite sum, it will
converge to a finite value.
Now, with that out of the way for review,

Bulgarian: 
просто умножаваме по r,
като всичко това тук е проговор.
Ако за теб не е преговор,
ти препоръчвам
да гледаш клиповете за
геометрични прогресии.
Интересното за това е, 
че сме доказали вече
в уроците за сума на геометрична
прогресия,
че частното, ако абсолютната
стойност на частното е по-малко от 1,
то тогава този ред е сходящ
(критерий на Даламбер).
И ако абсолютната стойност на r
е по-голяма ли равна на 1,
тогава редът е разходящ
(критерий на Даламбер).
Това е логично,
ние го доказахме,
но то също така е и логично, че
ако абсолютната стойност
на r е по-малко от 1,
то всеки член на реда ще е по-малък 
с коефициент, равен на частното,
той е произведение с
това частно,
и членовете на реда
ще стават по-малки и по-малки,
и е напълно логично,
че макар това да е
една безкрайна сума, тя ще клони 
към някаква крайна стойност.
Сега, след като преговорихме това,

Portuguese: 
está apenas multiplicando, 
apenas multiplicando por r,
e isso é tudo revisão.
Se não for revisão, recomendo
que assista aos vídeos 
sobre série geométrica.
O interessante
é que provamos
para nós mesmos, nos vídeos
sobre a série geométrica,
que se a razão,
se o valor absoluto
da razão é inferior a um,
então a série converge.
E se o valor absoluto de r
é maior do que ou igual a um,
então a série diverge.
E isso faz sentido,
provamos isso também,
mas também faz sentido
se o valor absoluto
de r é inferior a um
então cada termo seguinte
será diminuído
ou será multiplicado
por essa razão
e vai diminuir
até ser aproximar, mesmo
sendo uma soma infinita, ele vai
convergir para um valor finito.
Agora, saindo da revisão,

Thai: 
ลองมาทำสิ่งที่น่าสนใจกว่านี้ดีกว่า
สมมุติว่าเราอยากดู
เราอยากหาว่าอนุกรมแบบนี้
เริ่มที่ n เท่ากับ 5 ถึงอนันต์
อย่างเช่น n กำลัง 10
ตัวเศษโตขึ้นอย่างรวดเร็ว
n กำลัง 10 ส่วน n แฟคทอเรียล
และแฟคทอเรียล เรารู้ว่ามันโตเร็วมากๆ เช่นกัน
น่าจะโตเร็วกว่า
แม้แต่พหุนามดีกรีสูงอย่างนี้
หรือเทอมดีกรีสูงอย่างนี้
แต่เราจะพิสูจน์ได้อย่างไรว่ามันลู่เข้า?
เราใช้การทดสอบการลู่ออกได้
เพื่อแสดงว่ามันไม่ได้ลู่ออก
แต่เราจะพิสูจน์ได้อย่างไรว่ามันลู่เข้าจริงๆ?
บางทีเราต้องใช้สัญชาตญาณนิดหน่อยตรงนี้
ลองดูว่าเราหาอัตราส่วนร่วมได้ไหม
ลองดูว่ามันมีอัตราส่วนร่วมตรงนี้ไหม
ลองหาเทอมที่ n บวก 1
ซึ่งก็คือ n บวก 1 กำลัง 10

English: 
let's tackle something a
little bit more interesting.
Let's say that we want to look at a,
we want to figure out
whether a series like this,
so starting at N equals five to infinity
of, let's say, N to the tenth power.
The numerator is growing quickly.
N to the tenth power over N factorial,
and factorial we know also
grows very, very quickly,
probably it grows much faster than
even a high degree polynomial like this,
or a high degree term like this,
but how do we prove that it converges?
We could definitely the diverges test
to show that this does not diverge,
but how do we prove that
this actually converges?
And maybe we can use a little
bit of our intuition here.
Well, let's see if we can
come up with a common ratio.
So let's see if there's
a common ratio here.
So let's take the N plus oneth term,
which is going to be N
plus one to the tenth power

Portuguese: 
vamos enfrentar algo um
pouco mais interessante.
Suponha que queiramos olhar para,
descobrir se uma série como esta,
partindo de n igual a cinco ao infinito
de, digamos, n à décima potência.
O numerador está crescendo rapidamente.
n à décima potência sobre fatorial de n
e sabemos que fatoriais 
crescem muito rapidamente,
provavelmente, muito mais rápido
mesmo um polinômio de 
grau elevado como este,
ou um termo de alto grau como este,
mas como vamos provar que converge?
Poderíamos testar sua divergência
para mostrar que não diverge,
mas como vamos provar que
realmente converge?
E talvez possamos usar um pouco
de nossa intuição aqui.
Bem, vamos ver se 
chegamos a uma razão.
Então, vamos ver se há
uma razão aqui.
Então, vamos tomar o 
enésimo mais um termo,
que será n mais um elevado
a décima potência

Bulgarian: 
да видим нещо още по-интересно.
Да кажем, че искам
да разгледам,
искам да определя дали 
един ред като този,
за n от 5 до безкрайност
от n^10...
Числителят нараства бързо,
n^10 върху n! (n факториел),
а факториелът знаем, че
нараства много, много бързо,
вероятно нараства
много по-бързо
даже от полином на такава
висока степен,
или от член на толкова висока степен,
но как да докажем, че
това е сходящо?
Можем да използваме необходимото
условие за сходимост,
за да покажем, че 
този ред не е разходящ,
но как да докажем, че
той всъщност е сходящ?
Можем да използваме
подобна логика.
Да видим дали можем
да определим частното.
Да видим има ли тук частно.
Взимаме (n + 1)-члена,
който ще бъде (n + 1)^10

Korean: 
좀 더 흥미로운 것들을 배워봅시다
예를 들어 우리가
이러한 수열에 대해 알아보고 싶다고 합시다
n은 5부터 ∞까지 증가하고
분자는 n^10이라고 합시다
분자가 빠르게 커집니다
∑n^10/n!
그리고 계승 또한 매우 빠르게 커집니다
아마도 이러한 높은 차수의 다항식보다도
더욱 빨리 커질 것입니다
더욱 빨리 커질 것입니다
이것이 수렴한다는 것을 어떻게 증명할까요?
발산 판정법을 통해
발산하지 않는다는 것은 증명할 수 있지만
이것이 실제로 수렴한다는 것은 어떻게 증명할까요?
우리의 직감을 사용할 수 있습니다
일단 공비를 찾을 수 있는지 봅시다
일단 여기에 공비가 있는지 봅시다
일단 n+1번째 항을 보면
(n+1)^10/(n+1)!일 것이고

Thai: 
ส่วน n บวก 1 แฟคทอเรียล
แล้วหารด้วยเทอมที่ n
ลองหารมันด้วย n กำลัง 10 
ส่วน n แฟคทอเรียล
การหารด้วยเศษส่วน หรือหารด้วยอะไรก็ตาม
เทียบเท่ากับการคูณด้วยส่วนกลับของมัน
ลองคูณด้วยส่วนกลับ
คูณ n แฟคทอเรียลส่วน n กำลัง 10
นึกดู ที่ผมพยายามทำคืออันนี้ตรงนี้
ดูว่ามันมีอัตราส่วนร่วมไหม
ถ้าเราคิดเลขตรงนี้
n บวก 1 แฟคทอเรียล 
การคิดแฟคทอเรียลนั้นสนุก
อันนี้เท่ากับ n บวก 1 คูณ n แฟคทอเรียล
คูณ n แฟคทอเรียล
คุณอาจจะไม่คุ้นกับ
ลำดับการดำเนินการที่มีแฟคทอเรียล
แต่แฟคทอเรียลนี้ใช้กับ n นี่ตรงนี้อย่างเดียว
แล้วทำไมมันถึงมีประโยนชน์
เพราะ n แฟคทอเรียลนี้ตัดกับ n แฟคทอเรียลนี้

Bulgarian: 
върху (n + 1)! и разделяме това
на n-тия член,
значи разделяме на n^10/n!.
Когато делим на дроб,
или когато делим на нещо,
това е равносилно на
умножение по реципрочното му.
Така че неща просто
умножим по реципрочната стойност,
значи по n!/n^10.
Спомни си какво се 
опитвам да направя ето тук.
Търся дали имаме частно.
Сега ще използваме
малко алгебрични преобразувания,
(n + 1)!, сметките
с факториел са винаги забавни.
Това е равно на (n +1) по n!,
по n факториел.
Тук нямаме ред на действията,
когато работим с факториел.
Този факториел тук
се отнася само за това n.
Защо това е полезно?
Защото този n! се съкращава
с този n!,

Portuguese: 
sobre fatorial de n mais um e
divida pelo enésimo termo,
então vamos dividir por n à décima 
potência sobre fatorial de n.
Bem, dividir por uma fração, ou
por qualquer coisa
equivale a multiplicar por seu inverso,
Então vamos multiplicar pelo inverso,
assim, vezes fatorial de n
sobre n à decima potência.
Lembre-se de estou fazendo como fiz aqui.
Veja se há alguma razão
Bem, fazendo um pouco de álgebra
fatorial de n mais um
Equivale a n mais um 
vezes fatorial de n
Você não está acostumado a ver
ordem nas operações com fatoriais,
mas aplica-se apenas o
fatorial a n,
e por que é tão útil?
Porque este fatorial de n
cancela com este fatorial de n,

Korean: 
이것을 n번째 항으로 나눕시다
그래서 이걸 n^10/n!으로 나누면
나눌 때에는 그것의 역수를 곱하는 것과 같으므로
나눌 때에는 그것의 역수를 곱하는 것과 같으므로
역수를 곱해주면
(n+1)^10/(n+1) * n!/n^10입니다
제가 하려는 것은 위에서 했던 것과 같습니다
어떠한 공비가 있는지를 보는 것입니다
조금의 계산을 해보면
(n+1)!은 (n+1)*n!과 같습니다
(n+1)!은 (n+1)*n!과 같습니다
(n+1)!은 (n+1)*n!과 같습니다
계승이 있는 연산의 우선순위를 따지는 것은
익숙하지 않지만
이 계승은 여기에 n에만 적용됩니다
그리고 왜 이것이 도움이 되냐하면
이 n!은 저 n!과 상쇄되고

English: 
over N plus one factorial and
divide that by the Nth term,
so let's divide that by N to
the tenth over N factorial.
Well, dividing by a fraction,
or dividing by anything,
is equivalent to multiplying
by it's reciprocal,
So let's just multiply by the reciprocal,
so times N factorial over N to the tenth.
Remember all I'm trying to do
is exactly right over here.
See if there is some
type of a common ratio.
Well, if we do a little algebra here
N plus one factorial, and
factorial algebra is always fun.
This is the same thing as N
plus one times N factorial.
Times N factorial.
You're not use to seeing order
of operations with factorials,
but this factorial only applies
to this N right over here,
and why is that useful?
Because this N factorial
cancels with this N factorial,

Portuguese: 
e ficamos com n mais um
à décima potência
sobre n mais um vezes n elevado a dez
Sei o que está pensando.
"Ei, espere, olhe, isso não é
razão constante".
A razão entre termos consecutivos aqui,
quando tomei o enésimo mais um termo
dividido pelo enésimo termo
está variando, dependendo do meu n.
Esta razão, você
poderia dizer, é uma função de n
de modo que não parece muito útil,
mas eu diria:
"O.K., bem, com qualquer uma dessas séries
realmente nos importamos com
o comportamento de n se
ficar realmente, realmente grande
no limite, nossos ns tendem ao infinito."
Então, se observarmos o 
comportamento,
e, se este se aproxima a
valores reais se n tende a infinito,
bem, faria sentido
pensarmos como
o limite da nossa razão.
Então, vamos fazer isso.
Vamos tomar o limite de n
tendendo ao infinito dessa coisa.

English: 
and we're left with N plus
one to the tenth power
over N plus one times N to the tenth.
Times N to the tenth.
So I know what you're thinking.
"Hey, wait, look, this isn't
a fixed common ratio here."
The ratio between consecutive terms here,
when I took the N plus oneth
term divided by the Nth term,
it's changing depending on what my N is.
This ratio, I guess you
could say, is a function of N
so this doesn't seem too useful,
but what if I were to say,
"O.K., well look, with any of these series
"we really care about
the behavior as our N's
"get really, really, really large as
the limit, as our N's, go to infinity."
So what if we were to
look at the behavior here,
and if this is approaching some
actual values as N approaches infinity,
well, it would make conceptual sense
that we could kind of think of that as
the limit of our common ratio.
So let's do that.
Let's take the limit as N
approaches infinity of this thing.

Korean: 
그렇게 되면 (n+1)^10/(n+1)n^10이 남기 때문입니다
그렇게 되면 (n+1)^10/(n+1)n^10이 남기 때문입니다
그렇게 되면 (n+1)^10/(n+1)n^10이 남기 때문입니다
저는 여러분이 무슨 생각을 하는지 압니다
이 값은 고정된 공비가 아닙니다
n+1항을 n항으로 나누면
연속되는 두 항 사이의 비가
n값이 무엇인지에 따라서 달라집니다
여기서 공비는 n에 대한 함수라고 할 수 있습니다
그래서 이것은 많이 도움이 되어보이진 않지만
만약 제가
이러한 어떤 수열이든 간에
우리가 정말로 중요하게 생각하는 것은
n이 매우 많이 커져서
n이 ∞로 갈때라고 한다 해봅시다
만약 우리가 이 함수의 성질을 봤을 때
n이 무한대로 가면서 어떤 유한한 값에 수렴한다면
n이 무한대로 가면서 어떤 유한한 값에 수렴한다면
우리가 그것을 공비의 극한이라고
생각해도 개념적으로 타당합니다
생각해도 개념적으로 타당합니다
그러니까 그렇게 합시다
lim(n+1)^10/(n+1)n^10

Thai: 
แล้วเราเหลือ n บวก 1 กำลัง 10
ส่วน n บวก 1 คูณ n กำลัง 10
คูณ n กำลัง 10
ผมรู้ว่าคุณคิดอะไร
เฮ้ เดี๋ยวก่อน นี่ไม่ใช่อัตราส่วนร่วมคงที่นี่
อัตราส่วนระหว่างเทอมที่ติดกันตรงนี้
เมื่อผมนำเทอมที่ n บวก 1 
มาหารด้วยเทอมที่ n
มันจะเปลี่ยนค่าไปขึ้นอยู่กับว่า n ผมคืออะไร
อัตราส่วนนี้ คุณจะเรียกว่า
เป็นฟังก์ชันของ n ก็ได้
ตัวนี้จึงไม่ค่อยมีประโยชน์นัก
แต่ถ้าเกิดผมบอกว่า
โอเค ดูนะ สำหรับอนุกรมใดๆ
เราสนใจแค่พฤติกรรมเมื่อ n
มีค่ามากๆๆๆๆ
เป็นลิมิต เมื่อ n ของเราไปหาอนันต์
ถ้าเกิดเราดูพฤติกรรมตรงนี้
และถ้าตัวนี้เข้าใกล้
ค่าจริงเมื่อ n เข้าหาอนันต์
มันก็สมเหตุสมผล
ที่เราคิดถึงมันเป็น
ลิมิตของอัตราส่วนร่วม
ลองทำกันดู
ลองหาลิมิตเมื่อ n เข้าหาอนันต์ของตัวนี้

Bulgarian: 
и ни остава само (n + 1)^10
върху (n + 1) по n^10.
Знам какво си мислиш сега.
"Тук няма фиксирано частно."
Частното между съседните 
членове тук,
когато разделя (n + 1)-вия член
на n-тия член,
частното се променя
в зависимост от n.
Това частно, ако мога
да кажа така, е функция от n,
така че това не 
изглежда да върши работа.
Обаче, ако кажа:
За всеки от тези редове,
се интересуваме от поведението,
когато нашето n
стане много, много голямо,
когато n клони към безкрайност.
Ако разгледаме поведението тук,
ако това клони към някакви
действителни стойности,
когато n клони към безкрайност?
Тук изглежда логично,
че можем да разглеждаме
това като граница на частното.
Хайде да го направим.
Да намерим границата на това,
когато n клони към безкрайнсот.

Portuguese: 
Então não se esqueça,
Este não é apenas algum tipo de voodoo.
Vamos refletir
sobre o conceito de razão.
O que é a razão entre dois 
termos consecutivos?
O termo na posição n mais um, 
e o enésimo termo
para n realmente, realmente grande?
E o que vemos aqui
nem temos que abrir a expressão
será n elevado a dez
mais um monte de outras coisas.
Será um polinômio de décimo grau,
e este outro aqui
na verdade, você pode descobrir,
será n elevado a onze 
mais n elevado a dez
assim, no limite para n
tendendo ao infinito,
o denominador 
cresce mais rápido
Podemos dividir o numerador
e o denominador por n elevado a dez,
na verdade, o numerador e o
denominador por n elevado a 11,
e tudo aqui
tende a zero,
e este será um,
e assim o limite, bem aqui,

English: 
So remember what we're doing here.
This isn't just kind of some voodoo.
Conceptually, let me copy and paste this.
Conceptually, we're just trying to think
about what is the common ratio approach?
What is the ratio between
consecutive terms?
The N plus one term and the Nth term
as N gets really, really,
really, really large?
And what we see here is
we have this up here,
we don't even have to multiply it out,
this is going to be N to the tenth
plus a bunch of other stuff.
It's going to be a
tenth degree polynomial,
and this down here is going to be,
actually you can figure it out,
this is going to be N to the
11th plus N to the tenth power,
so the limit as N approaches infinity,
well this denominator is going
to grow faster than that.
You could divide the numerator
and the denominator by N to the tenth,
actually the numerator and the
denominator by N to the 11th,
and everything up here
is going to go to zero,
and this is going to be one,
and so the limit, right over here,

Bulgarian: 
Спомни си какво
правим тук.
Това не е някаква вуду магия.
Сега ще копирам
и ще поставя това.
Теоретично, ние се опитваме
да оценим
към какво се доближава
нашето частно.
Колко е частното
между съседните членове?
Между члена (n + 1) и n-тия член,
когато n става
много, много голямо?
Виждаме ето това тук,
дори няма нужда
да умножаваме,
това ще бъде n^10
плюс всичко това.
Това е полином от 10-та степен,
а това тук ще бъде...
всъщност може да го намериш –
това ще бъде n^11 плюс n^10.
Така че границата, когато
n клони към безкрайност,
тогава този знаменател
ще нараства по-бързо от това.
Можем да разделим
числителя
и знаменателя на n^10,
всъщност числителя
и знаменателя на n^11,
и всичко тук горе
ще стане нула,
това ще стане 1,
тогава границата ето тук,

Thai: 
นึกถึงสิ่งที่เราทำอยู่ตรงนี้
นี่ไม่ใช่เวทมนตร์อะไร
โดยหลักการแล้ว ขอผมลอกและวางตัวนี้นะ
โดยหลักการแล้ว เราแค่คิด
ว่าอัตราส่วนร่วมเข้าหาอะไร?
อัตราส่วนระหว่างเทอมที่ติดกันเป็นเท่าใด?
เทอมที่ n บวก 1 กับเทอมที่ n
เมื่อ n มีค่ามากๆๆๆ เป็นเท่าใด?
และสิ่งที่เราเห็นตรงนี้ คือเรามีบนนี้
เราไม่ต้องคูณมันออกมาด้วยซ้ำ
อันนี้จะเท่ากับ n กำลัง 10
บวกตัวอื่นๆ
มันจะเท่าากับพหุนามดีกรี 10
และตัวนี้ข้างล่างจะเป็น
คุณหามันได้
อันนี้จะเป็น n กำลัง 11 บวก n กำลัง 10
ลิมิตเมื่อ n เข้าหาอนันต์
ตัวส่วนจะโตเร็วกว่าตัวเศษ
คุณหารตัวเศษ
และตัวส่วนด้วย n กำลัง 10 ได้
หารตัวเศษและตัวส่วนด้วย n กำลัง 11 ดีกว่า
แล้วทุกอย่างตรงนี้จะไปหา 0
อันนี้จะเป็น 1
แล้วลิมิต ตรงนี้

Korean: 
우리가 무엇을 하려는지 생각해봅시다
이것은 이상한 주술같은 것이 아닙니다
이것을 옮겨 적겠습니다
우리는 공비가 어떤 값에 수렴하는지
생각해보려고 하는 것입니다
두 항 사이의 비율이 무엇인지를 생각해보려는 것입니다
n이 매우 많이 커지면
n항과 n+1항 사이의 공비가 어떻게 되는지 생각합시다
우리가 분자에서 알 수 있는 것은
이것을 곱할 필요도 없이
n^10과 다른 항들의 합이라는 것을 알 수 있습니다
n^10과 다른 항들의 합이라는 것을 알 수 있습니다
최대 차수가 10인 다항식일 것입니다
분모는
실제로 계산해보면
n^11 + n^10일 것입니다
n이 무한대로 발산하게 되면
분모가 분자보다 더 빠르게 커지게 됩니다
분자와 분모를 n^10로 나누면
분자와 분모를 n^10로 나누면
아니 n^11로 나누면
분자는 0으로 수렴할 것이고
분모는 1으로 수렴할 것입니다
그러니까 이 lim값은 0이 될 것입니다

Thai: 
ลิมิตนี้จะเท่ากับ --
ลิมิตนี้จะเท่ากับ 0
เมื่อ n เข้าหาอนันต์ อัตราส่วน
ระหว่างเทอมที่ติดกันเข้าหา 0
อันนี้ดูเหมือนว่า ถ้าเราใช้เหตุผลเรื่องอัตรา
ส่วนร่วมบนนี้ตอนเราดูอนุกรมเรขาคณิต
อันนี้ไม่ใช่อนุกรมเรขาคณิตแน่นอน
แต่เราบอกว่า ดูสิ เมื่อ n
มีค่ามากๆๆๆๆ อัตราส่วนระหว่าง
เทอมที่ติดกันยิ่งเล็กลง เล็กลง และเล็กลง
เฮ้ บางทีเราอาจสรุปได้เหมือนกัน
บางที เราอาจสรุปได้ว่า
อนุกรมนี้จริงๆ แล้วลู่เข้า
อนุกรมนี้จริงๆ แล้วลู่เข้า n แฟคทอเรียลลู่เข้า
เราสรุปอย่างนั้นได้ไหม?
คำตอบคือใช่ เราทำได้
และสิ่งที่อนุญาตให้เราทำได้ 
คือการทดสอบอัตราส่วน
คือการทดสอบอัตราส่วน
ขอผมเขียนมันลงไปนะ
การทดสอบอัตราส่วน

English: 
this limit is going to be equal to...
This limit is going to be equal to zero.
As N approaches infinity one way the ratio
between consecutive terms approaches zero.
So this seems, if we use
the logic from our common
ratio up here when we
looked at geometric series,
this is clearly not a geometric series,
but we would say, "Well, look, as N gets
really, really, really
large the ratio between
"consecutive terms gets smaller
and smaller and smaller.
"Hey, maybe we could
make the same conclusion.
"Maybe we could make
the same conclusion that
"therefore this series
actually converges."
The series actually converges,
so N factorial converges.
So can we make this statement?
And the answer is yes we can,
and what allows us to
do it is the ratio test.
Is the ratio test.
Let me write that down.
The ratio test.

Bulgarian: 
тази граница ще бъде равна на...
Тази граница ще бъде
равна на нула.
Когато n клони към безкрайност,
частното между съседните 
членове клони към нула.
И тук изглежда, ако 
използваме логиката от
частното тук, когато
разглеждахме геометрична прогресия,
това очевидно не е 
геометрична прогресия,
но можем да кажем:
"Когато n стане много, много голямо,
частното между съседните членове
става все по-малко и по-малко."
Може би можем да направим
същия извод, че
следователно този ред
е практически сходящ.
Редът практически е сходящ, така че 
редът с общ член n^10/n! е сходящ.
Можем ли да твърдим това?
Отговорът е да, можем,
и можем благодарение на граничната
форма на критерия на Даламбер.
Ще го запиша.
Граничната форма на критерия
на Даламбер.

Portuguese: 
este limite será igual a ...
Este limite será igual a zero.
Com n tendendo a infinito, a razão
entre dois termos consecutivos 
tende a zero.
Portanto, se usarmos
a lógica da nossa
razão aqui quando 
olhamos a série geométrica,
que não é uma série geométrica,
mas gostaríamos de dizer: 
"Bem, conforme n
fica realmente, realmente
grande a razão entre
termos consecutivos 
fica cada vez menor.
"Ei, talvez pudéssemos
chegar à mesma conclusão
de que esta série
na verdade, converge. "
A série converge na verdade,
assim como o fatorial de n converge.
Então, podemos fazer essa afirmação?
E a resposta é sim, podemos,
e que nos permite
fazê-lo é o teste da razão.
O teste da razão.

Korean: 
그러니까 이 lim값은 0이 될 것입니다
그러니까 이 lim값은 0이 될 것입니다
n이 무한대로 가면서 공비는 0이 될 것입니다
n이 무한대로 가면서 공비는 0이 될 것입니다
우리가 위에서 봤던 기하급수의
공비에 대한 논리를 보면
이것은 물론 기하급수가 아니긴 하지만
n이 엄청나게 커지면서
연속하는 두 항 사이의 비율이
점점 작아집니다
우리는 같은 결론을 내릴 수 있을까요?
이 함수가 결국 수렴한다는
같은 결론을 내릴 수 있을까요?
이 함수는 수렴한다는 문장이 참일까요?
이 함수는 수렴한다는 문장이 참일까요?
정답은 참입니다
이것을 성립하게 해주는 것은
비판정법입니다
여기에 적겠습니다
비판정법(Ratio Test)

Korean: 
비판정법이 우리에게 알려주는 것은
만약 무한급수가 하나 있고
만약 무한급수가 하나 있고
n은 k부터 무한대까지라고 하면
lim|An+1/An|을 봅시다
그리고 lim|An+1/An|을 보면
위에서는 절댓값을 취해주지 않았지만
위에서는 절댓값을 취해주지 않았지만
위에서는 모든 항이 양수였으므로
절댓값과 원래 값이 같으므로
절댓값을 취해줘도 됩니다
절댓값을 취해줘도 됩니다
만약 이것이 L에 수렴한다고 하면
만약 이것이 L에 수렴한다고 하면
이 값 L은 N이 계속해서 커졌을 때의
연속된 두 항 사이의 비율이 될 것이고
만약 L이 1보다 작다면
참고로 위의 상황에서도 L이 1보다 작을 때였고

English: 
And all the ratio test tells us is
if we have an infinite series...
If we have an infinite series,
and I can say from N equals K to infinity,
and if we take the limit
as N approaches infinity,
of the absolute value
of A to the N plus one
over A to the N, and over here I didn't
take the absolute value of that limit,
but we absolutely could have taken
the absolute value of that limit
if all of these terms right
over here were positive,
so the absolute value would
be the same thing as itself.
If this approaches some limit,
if this approaches some limit,
so once again, this limit is
what is going to be our ratio
between consecutive terms
as N gets larger and larger,
the ratio tells us if L is less than one,
which was the situation here,
L is clearly less than one,

Thai: 
และการทดสอบอัตราส่วนบอกเราว่า
ถ้าเรามีอนุกรมอนันต์ --
ถ้าเรามีอนุกรมอนันต์
และผมบอกได้ว่า จาก n เท่ากับ k ถึงอนันต์
และถ้าเราให้ลิมิตเมื่อ n เข้าหาอนันต์
ค่าสัมบูรณ์ของ a ห้อย n บวก 1
ส่วน a ห้อย n ตรงนี้ ผมไม่ได้
หาค่าสัมบูรณ์ของลิมิตนั้น
แต่เราหา
ค่าสัมบูรณ์ของลิมิตนั้นได้
ถ้าทุกเทอมตรงนี้เป็นบวก
ค่าสัมบูรณ์จะเท่ากับตัวมันเอง
ถ้าตัวนี้เข้าหาลิมิต
ถ้าค่านี้เข้าหาลิมิตค่าหนึ่ง
ย้ำอีกครั้ง ลิมิตนี้คือค่าอัตราส่วนของเรา
ระหว่างเทอมติดกันเมื่อ n มากขึ้น มากขึ้น
การทดสอบอัตราส่วนบอกเราว่า 
ถ้า L น้อยกว่า 1
ซึ่งก็คือกรณีตรงนี้ L น้อยกว่า 1 ชัดเจน

Portuguese: 
E o teste da razão nos diz que
se temos uma série infinita ...
e sei disso pois n varia de k ao infinito,
se tomarmos o limite
para n tendendo ao infinito,
do valor absoluto
do termo na posição n mais um
sobre o termo na posição n, e aqui ainda
não havíamos feito o limite
mas poderíamos ter tomado
o valor absoluto do limite
se todos esses termos à direita
aqui fossem positivos,
o valor absoluto 
será positivo.
Se este se aproxima de um limite,
então mais uma vez, ele será a nossa razão
entre termos consecutivos
para valores muito grandes de n
Se a razão L é menor que um,
que era a situação aqui,
L é claramente menor do que um,

Bulgarian: 
Той ни казва, че
ако имаме безкраен числов ред,
за n от k до безкрайност,
и ако намерим границата,
когато n клони към безкрайност,
границата на абсолютната
стойност на а^(n +1)
върху а^n, като тук не взимам
абсолютната стойност на границата,
но напълно можем да вземем
абсолютна стойност на границата,
абсолютната стойност
на тази граница,
ако всички тези членове ето тук
са положителни,
абсолютната стойност на
границата е равна на себе си.
Ако това клони към 
някаква граница,
отново, тази граница
е нашето частно
между последователните членове,
когато n става все по-голямо,
частното ни казва, че ако
L е по-малко от 1,
какъвто е случаят тук, L 
определено е по-малко от 1,

Portuguese: 
então a série converge.
Algumas definições dirão que 
é absolutamente convergente,
converge absolutamente,
o valor absoluto
das séries convergem,
o que também significa
que a própria série converge.
Se L é maior do que um
então, a série diverge.
e, se L é igual a um, é
inconclusiva, não sabemos.
Teríamos que fazer algum outro teste
para provar se ela converge ou diverge.
Então essa é a essência do teste da razão.
Determinar o valor absoluto da razão
entre termos consecutivos,
tomar o limite para n tendendo a infinito,
e, se aproximar de um limite real,
e for menor do que um, 
então a série converge,
e é baseado na mesma 
ideia fundamental
que vimos com a razão da série geométrica.
[Legendado por: Tatiana F. D'Addio]

English: 
then the series converges.
Some definitions will say
it's absolutely convergent,
it converges absolutely,
the absolute value
of the series converges,
but then that also means
that the series itself converges.
If L is greater than one
then series diverges,
Series diverges.
and if L is equal to one it's
inconclusive, we don't know.
We would have to do some other test
to prove whether it converges or diverges.
So that's the essence of the ratio test.
Find the absolute value of the ratio
between consecutive terms,
take the limit as N approaches infinity,
if that approaches an actual limit,
and that limit is less than
one, then the series converges,
and it's really based on
the same fundamental idea
that we saw with the common
ratio of geometric series.

Korean: 
그렇다면 수열이 수렴합니다
어떤 정의에서는 절대수렴한다고
즉 수열의 절댓값이 수렴한다고 하는데
이것은 곧 수열 자체가 수렴한다는 것을
의미합니다
만약 L이 1보다 크다면
수열은 발산하게 됩니다
그리고 만약 L이 1이라면 모릅니다
다른 판정법을 해봐서
수렴하는지 발산하는지 증명해봐야 합니다
이것이 비판정법의 핵심입니다
연속하는 두 항 사이의 비율의
절댓값을 찾아서
n이 무한대에 한없이 가까워질 때
그것이 어떤 값에 수렴하며
그 값이 1보다 작으면 무한급수가 수렴한다는 것입니다
이것은 모두 기하급수의 공비에서 본
기본적인 아이디어에 기반을 두고 있습니다
커넥트 번역 봉사단 | 한승주

Thai: 
อนุกรมนี้ก็จะลู่เข้า
นิยามบางตัวจะบอกว่า มันลู่เข้าโดยสัมบูรณ์
มันลู่เข้าโดยสัมบูรณ์ ค่าสัมบูรณ์
ของลำดับจะลู่เข้า แต่นั่นก็หมายความว่า
อนุกรมเองลู่เข้าด้วย
ถ้า L มากกว่า 1 แล้วอนุกรมจะลู่ออก
อนุกรมลู่ออก
และถ้า L เท่ากับ 1 มันจะสรุปไม่ได้ เราไม่รู้
เราต้องทำการทดสอบอื่น
เพื่อพิสูจน์ว่ามันลู่เข้าหรือลู่ออก
นั่นคือสาระของการทดสอบอัตราส่วน
หาค่าสัมบูรณ์ของอัตราส่วน
ระหว่างเทอมที่ติดกัน
หาลิมิตเมื่อ n เข้าหาอนันต์
ถ้ามันเข้าใกล้ลิมิตนั้น
และค่าลิมิตนั้นน้อยกว่า 1 อนุกรมก็จะลู่เข้า
และมันมาจากแนวคิดพื้นฐานเดียวกัน
กับที่เราเห็นอัตราส่วนร่วมในอนุกรมเรขาคณิต

Bulgarian: 
тогава редът е сходящ.
Според някои дефиниции,
той е абсолютно сходящ,
редът от абсолютните стойности на членовете му 
е сходящ, но това означава също,
че самият ред е сходящ.
Ако L е по-голямо от 1, тогава
редът е разходящ.
Ако L е равно на 1, тогава не можем да заключим,
не знаем дали редът е сходящ или разходящ.
Трябва да използваме
друг критерий,
за да докажем дали е
сходящ или разходящ.
Това е смисълът на граничната
форма на критерия на Даламбер.
Да намерим абсолютната
стойност на частното
между съседните членове,
намираме границата, когато
n клони към безкрайност,
и ако това клони към
действителна граница,
и тази граница е по-малко от 1,
тогава редът е сходящ,
като това почива на 
същия основен принцип,
който се отнася за частното на геометричните 
прогресии, където се прилага критерият на Даламбер.
