
English: 
I made a claim that for
this sequence-- and this
was in a previous video--
that for this sequence
right over here that can
be defined explicitly
in this way, that the
limit of the sequence--
and so I can write this as
negative 1 to the n plus 1
over n.
That's one way of defining
our sequence explicitly--
the limit of this as
n approaches infinity
is equal to 0.
And it seems that way.
As n gets larger and
larger and larger,
even though the numerator
oscillates between negative 1
and 1, it seems like
it will get smaller
and smaller and smaller.
But I didn't prove
it, and that's
what I want to do in this video.
In order to prove it, this is
going to be true if and only
if for any epsilon
greater than 0,
there is a capital
M greater than 0

Portuguese: 
Já afirmei em um vídeo anterior
que para esta sequência,
explicitamente expressa desta forma,
o limite desta sequência...posso escrevê-la
como menos 1 elevado à n mais 1
sobre n.
Esta é uma forma de definir nossa função.
O limite conforme n se aproxima ao infinito
é igual a 0.
E faz sentido.
Conforme n cresce mais e mais,
ainda que o numerador alterne entre menos 1
e 1, ele vai ficando cada vez menor
e menor, e menor.
Mas não provei isto,
e é o que quero fazer neste vídeo.
Isto apenas será verdade somente se
para qualquer épsilon maior que 0,
exista um M maior que 0, tal qual

Bulgarian: 
 
В предишното видео заявих
твърдение, че тази редица
ето тук може да бъде определена чрез явно задаване
по начин, при който границата на редицата --
мога да я напиша като минус 1 на степен n плюс 1
върху n.
Това е единият начин за определяне на дадена редица явно --
границата на това, когато n клони към безкрайност,
е равна на 0.
Като тя изглежда по този начин.
n става все по-голямо и по-голямо, и по-голямо,
въпреки че числителят се колебае между минус 1
и 1. Изглежда че той става все по-малък
и по-малък и по-малък.
Но не съм го доказал, като точно това
искам да направя сега във видеото.
За да го докажа, това ще бъде вярно ако и само
ако за всяко епсилон по-голямо от 0,
има главно М, което е по-голямо от 0,

Korean: 
지난 번 영상에서 다뤘던
이 수열은
여기 적힌 것 처럼
정의할 수 있었죠
{(-1)^(n+1)}/n이라는 형태로
이 수열을 나타낼 수 있고
n이 무한대로 갈 때
이 수열의 극한값은 0이라고
했었습니다
맞는 이야기죠
분모 n은 계속 커지는데
분자는 -1에서 1 사이를 
왔다갔다 하니까
수열 자체는
점점 작아질 겁니다
증명은 하지 않았었고
하고 싶지는 않지만
증명을 하려면
"0보다 큰 모든 입실론(ε)에 대해
M>0일 때

Thai: 
 
ผมได้อ้างไปว่า สำหรับลำดับนี้ -- อันนี้
คือวิดีโอที่แล้ว -- ว่าสำหรับลำดับนี่
ตรงนี้ที่เขียนนิยามได้โดยตรง
แบบนี้ ลิมิตของลำดับ --
ผมเเขียนอันนี้ได้เป็นลบ 1 กำลัง n บวก 1
ส่วน n
นั่นคือวิธีกำหนดลำดับโดยตรงวิธีหนึ่ง --
ลิมิตของตัวนี้เมื่อ n เข้าหาอนันต์
เท่ากับ 0
และมันดูเป็นเช่นนั้น
เมื่อ n โตขึ้น โตขึ้น และโตขึ้น
ถึงแม้ว่าตัวเศษจะแกว่งไปมาระหว่างลบ 1
กับ 1 มันดูเหมือน่ว่า มันจะเล็กลง
เล็กลง และเล็กลง
แต่ผมไม่ได้พิสูจน์ นั่น
คือสิ่งที่ผมอยากทำในวิดีโอนี้
เวลาพิสูจน์ อันนี้เป็นจริงก็ต่อเมื่อ
สำหรับเอปซิลอนที่มากกว่า 0 ใดๆ
จะมี M ใหญ่มากกว่า 0

Thai: 
โดยที่ถ้า n เล็ก ถ้าหมายเลขของเรามากกว่า
M ใหญ่ แล้วเทอมที่ n ในลำดับ
จะอยู่ในช่วงเอปซิลอนของลิมิต 
คือห่างจาก 0 ไม่เกินเอปซิลอน
 
แล้วมันหมายความว่าอะไร?
มันบอกว่า เฮ้ -- ลิมิตของเราคือ 0
ขอผมใช้สีใหม่นะ
ลิมิตของเราตรงนี้คือ 0
นั่นคือลิมิตของเรา
ลิมิตตรงนี้ก็คือ --
เราบอกว่าลำดับจะลู่เข้าหา 0
สิ่งที่เราบอกคือว่า ให้เอปซิลอนรอบ 0 เรามา
สมมุติว่าค่านี่ตรงนี้คือ 0 บวกเอปซิลอน
นั่นคือ 0 บวกเอปซิลอน
วิธีที่ผมวาดมันตรงนี้ 
ดูเหมือนว่าเอปซืลอนจะเป็น 0.5
นี่ก็คือ 0 ลบเอปซิลอน
ขอผมเขียนให้สวยหน่อย
นี่ก็คือ 0 ลบเอปซิลอน
นี่ก็คือลบเอปซิอลน, 0 ลบเอปซิลอน, 
0 บวกเอปซิลอน
ลิมิตของเราในกรณีนี้ 
หรือที่เราอ้างว่าเป็นลิมิต คือ 0
ทีนี้ อันนี้กำลังบอกว่า สำหรับเอปซิลอนใดๆ

English: 
such that if lowercase n, if our
index is greater than capital
M, then the nth
term in our sequence
is going to be within epsilon of
our limit, within epsilon of 0.
So what does that say?
That says, hey, give
me-- our limit is 0.
Let me do this in a new color.
So our limit right
over here is 0.
That's our limit.
So our limit right
over here is--
we're saying the sequence
is converging to 0.
What we're saying is, give
us an epsilon around 0.
So let's say that this right
over here is 0 plus epsilon.
That is 0 plus epsilon.
The way I've drawn it here
looks like epsilon would be 0.5.
This would be 0 minus epsilon.
Let me make it a
little bit neater.
So this would be
zero minus epsilon.
So this is negative epsilon, 0
minus epsilon, 0 plus epsilon.
Our limit in this case, or
our claim of a limit, is 0.
Now, this is saying
for any epsilon,

Bulgarian: 
При което ако малко n, ако индексът е по-голям от главно М,
тогава n-тият член от редицата
ще бъде в обхвата на епсилон за границата, в рамките на епсилон от 0.
 
Какво означава това?
Това означава, че -- границата ни е 0.
Нека го напиша с нов цвят.
Границата ни тук е 0.
Това е границата.
Границата тук е --
казваме, че редицата е сходяща към 0.
Казваме, че ако ни е дадено епсилон около 0.
Нека кажем, че това тук е 0 плюс епсилон.
Това е 0 плюс епсилон.
Начинът, по който го начертахме тук, изглежда че епсилон ще бъде 0,5.
Това ще бъде 0 минус епсилон.
Нека го начертая малко по-ясно.
Това ще бъде 0 минус епсилон.
Имаме отрицателен епсилон, 0 минус епсилон, 0 плюс епсилон.
Границата в този случай или твърдението ни за границата е 0.
Това гласи, че за всяко епсилон,

Korean: 
n>M이면
이 수열의 n번째 항과 
극한값, 즉 0 사이의 거리는
ε보다 작다"라는
필요충분 조건이 필요합니다
무슨 뜻일까요?
지금 이 수열의
극한값이 0이죠
극한값이 0이고요
극한값이 0이니까
이 수열은 0으로
수렴합니다
0보다 조금 큰
ε이 있다고 가정합시다
여기 쯤이 0+ε이 되겠죠
그려놓고 보니
ε이 0.5정도 돼 보이네요
여기가 0-ε입니다
좀 더 깔끔하게 그려볼게요
여기가 0-ε입니다
각각 0-ε, 0+ε입니다
지금은 극한값이 0이죠
이제 모든 ε에 대해서

Portuguese: 
se nosso índice n for maior que nosso M,
então o enésimo termo da nossa sequência
estará a menos de épsilon do limite,
dentro de épsilon de distância de 0.
Então o quê isto quer dizer?
Nosso limite aqui é 0.
Deixe-me mudar de cor.
Então o nosso limite aqui é 0.
Este é o nosso limite.
Então se nosso limite aqui é --
estamos dizendo que a sequência converge a 0.
Estamos dizendo que, dado um épsilon ao redor de 0.
Digamos que isto seja 0 mais épsilon.
Isto é 0 mais épsilon.
Do jeito que desenhei parece que épsilon é 0,5.
E isto seria 0 menos épsilon.
Vou deixar isso mais limpo.
Então isto seria 0 menos épsilon.
Isto é 0 menos épsilon, 0 mais épsilon.
Alegamos que o limite neste caso é 0.
Isto é dizer que para qualquer épsilon, precisamos

Portuguese: 
encontrar um M tal que se n for maior que M
a distância entre a sequência e nosso limite
será menor que épsilon.
Se a distância entre a nossa sequência e o nosso limite
for menor que épsilon, isso significa que o valor
da nossa sequência para um dado n
estará entre estes dois limites.
Estaria dentro desta área
que estou preenchendo, a partir de certo n.
Então se escolhesse um n aqui,
parece que qualquer n maior que isso
este será o caso, estaremos
entre estes limites.
Mas como provamos isto?
Bom, pensemos sobre o que
precisa acontecer para isto ser verdade.
O que precisa acontecer
para o valor absoluto de an menos 0
ser menor que épsilon?
Bem, isto é outra forma de dizer que
o valor absoluto de an tem que ser menor que épsilon.
E an é apenas este negócio aqui,
então é uma forma de dizer que o valor absoluto

Korean: 
n보다 작은 M을
찾을 건데,
이 때 이 수열과
그 극한값의 거리는
ε보다 작아야 합니다
수열과 극한값 간의 거리가
ε보다 작다는 뜻은
n이 어떤 값이든 간에
수열의 절댓값이
ε보다 작다는 거겠죠
이 색칠 된 영역 안에
존재해야 합니다
n이 어떤 값이든 말이죠
n을 여기 쯤으로
잡았다면
그 다음 항들도
이 영역 안에 포함되는 걸
알 수 있습니다
이걸 어떻게 증명할까요?
일단, 어떻게 해야
이 부분이 참이 되는지
생각해봅시다
어떻게 해야
(a_n)-0의 절댓값이
ε보다 작아질까요?
다른 식으로 표현하면
|(a_n)|<ε이죠
a_n은 여기 이 수열이니까
또 다르게 표현하면

English: 
we need to find an M such
that if n is greater than M,
the distance between our
sequence and our limit
is going to be
less than epsilon.
So if the distance between
our sequence and our limit
is less than epsilon,
that means that the value
of our sequence for
a given n is going
to be within these two bounds.
It's got to be in this
range right over here
that I'm shading
above a certain n.
So if I pick an n
right over here,
it looks like anything
larger than that
is going to be the case that
we're going to be within those
bounds.
But how do we prove it?
Well, let's just
think about what
needs to happen for
this to be true.
So what needs to be happen
for a sub n minus 0,
the absolute value
of a sub n minus 0,
what needs to be true for
this to be less than epsilon?
Well, this is
another way of saying
that the absolute value of a sub
n has to be less than epsilon.
And a sub n is just this
business right here,
so it's another way of saying
that the absolute value

Thai: 
เราต้องหา M โดยที่ ถ้า n มากกว่า M
ระยะระหว่างลำดับกับลิมิตของเรา
จะน้อยกว่าเอปซิลอน
ถ้าระยะระหว่างลำดับของเรากับลิมิต
น้อยกว่าเอปซิลอน นั่นหมายความว่าค่า
ของลำดับเราสำหรับ n ที่กำหนด
จะอยู่ภายในขอบสองตัวนี้
มันจะอยู่ในช่วงนี่ตรงนี้
ที่ผมแรเงาเหนือ n
ถ้าผมเลือก n ตรงนี้
มันดูเหมือนว่าอะไรก็ตามที่มากกว่านั้น
จะได้ว่า เราอยู่ในขอบเหล่านั้น
 
แต่เราจะพิสูจน์ได้อย่างไร?
ลองคิดดูว่า
ต้องมีอะไรจึงจะทำให้ประโยคนี้เป็นจริง
ต้องเกิดอะไรขึ้นเพื่อให้ a ห้อย n ลบ 0
ค่าสัมบูรณ์ของ a ห้อย n ลบ 0
อะไรต้องเป็นจริง 
เพื่อให้ตัวนี้น้อยกว่าเอปซิลอน?
พูดอีกอย่างคือว่า
ค่าสัมบูรณ์ของ a ห้อย n 
ต้องน้อยกว่าเอปซิลอน
a ห้อย n ก็คือพจน์นี่ตรงนี้
มันก็เหมือนกับบอกว่า ค่าสัมบูรณ์

Bulgarian: 
трябва да намерим такова М, при което ако n е по-голямо от М,
разстоянието между редицата и границата
ще бъде по-малко от епсилон.
Ако разстоянието между редицата и границата
е по-малко от епсилон, това означава, че стойността
на редицата за дадено n ще бъде
в рамките на тези две граници.
Тя трябва да е в този обхват ето тук,
който защриховам над определено n.
Следователно ако избера n да е ето тук,
изглежда, че всичко по-голямо от това
ще бъде случай, в който ще бъдем в рамките на
тези граници.
Но как ще го докажем?
Нека просто помислим, какво
трябва да се случи, за да бъде това вярно.
Какво трябва да е вярно, за да бъде а с индекс n минус 0,
абсолютната стойност на а с индекс n минус 0,
да бъде по-малко от епсилон?
Това е друг начин да кажем,
че абсолютната стойност на а с индекс n трябва да бъде по-малка от епсилон.
а с индекс n е просто това тук,
така че това е друг начин да кажем, 
че абсолютната стойност

Thai: 
ของลบ 1 กำลัง n บวก 1 ส่วน n
ต้องน้อยกว่าเอปซิลอน ซึ่งก็คือการบอกว่า
เนื่องจากลบ 1 กำลัง n บวก 1 นี้
ตัวเศษนี้แค่สลับค่าลบ
กับค่าบวกของ 1 ส่วน n
ถ้าคุณหาค่าสัมบูรณ์ของมัน
มันจะออกมาเป็นบวกเสมอ
ค่านี้จึงเท่ากับ 1 ส่วน n
เมื่อค่าสัมบูรณ์ของ 1 ส่วน n 
ต้องน้อยกว่าเอปซิลอน
ทีนี้ n จะเป็นบวกเสมอ
n เริ่มที่ 1 และไปถึงอนันต์
ค่านี้จึงต้องเป็นบวกเสมอ
อันนี้บอกว่า เท่ากับ 1 ส่วน n
ต้องน้อยกว่าเอปซิลอนเพื่อให้ตัวนี้
เป็นจริง
และตอนนี้เราหาส่วนกลับของทั้งสองด้านได้
และถ้าคุณหาส่วนกลับทั้งสองข้างของอสมการ
คุณจะได้ว่า n -- ถ้าคุณหา
ส่วนกลับทั้งสองข้างของอสมการ
คุณจะสลับอสมการ
เพื่อให้อันนี้เป็นจริง n ต้องมากกว่า 
1 ส่วนเอปซิลอน
และเราได้พิสูจน์มันไปแล้ว

Korean: 
{(-1)^(n+1)}/n의 절댓값이
ε보다 작아야 합니다
분자인 (-1)^(n+1)때문에
전체 값 자체는 1/n과 -(1/n)로
왔다갔다 할겁니다
하지만 절댓값을 취하면
항상 양수가 되겠죠
따라서 좌변은 1/n과 같고
|1/n|<ε이어야 합니다
n은 항상 양수입니다
1부터 무한대까지 가능하죠
따라서 항상 |1/n|>0입니다
그러니 1/n<ε이어야
|a_n-0|<ε이 참이 된다고도 
볼 수 있습니다
이제 양 변에
역수를 취합니다
부등식에서 역수를 취하면
부등호 방향도 바뀌어야죠
따라서 n>1/ε이 됩니다
이건 이미 증명됐죠

English: 
of negative 1 to
the n plus 1 over n
has to be less than epsilon,
which is another way of saying,
because this negative
1 to the n plus 1,
this numerator just swaps
us between a negative
and a positive
version of 1 over n.
But if you take the
absolute value of it,
this is always just
going to be positive.
So this is the same
thing as 1 over n,
as the absolute value of 1 over
n has to be less than epsilon.
Now, n is always
going to be positive.
n starts at 1 and
goes to infinity.
So this value is always
going to be positive.
So this is saying the
same thing that 1 over n
has to be less than epsilon
in order for this stuff
to be true.
And now we can take the
reciprocal of both sides.
And if you take the reciprocal
of both sides of an inequality,
you would have that
n-- if you take
the reciprocal of both
sides of an inequality,
you swap the inequality.
So for this to be true, n has to
be greater than 1 over epsilon.
And we essentially
have proven it now.

Bulgarian: 
на минус 1 на степен n плюс 1 върху n
трябва да бъде по-малка от епсилон, което е друг начин да кажем,
защото това минус 1 на степен n плюс 1,
този числител просто се сменя от отрицателна
и положителна версия на 1 върху n.
Но ако изчислиш абсолютната стойност от това,
то винаги ще бъде положително.
Това е същото като 1 върху n,
като абсолютната стойност от 1 върху n трябва да е по-малка от епсилон.
n ще бъде винаги положително.
n започва от 1 и отива до безкрайност.
Тази стойност ще бъде винаги положителна.
Това означава същото като 1 върху n
трябва да е по-малко от епсилон, за да може това нещо тук
да бъде вярно.
Сега пишем реципрочното от двете страни.
Ако вземем реципрочното от двете страни за едно неравенство,
ще имаме n -- ако вземеш
реципрочното от двете страни на едно неравенство,
разменяш знака му.
За да бъде това вярно, n трябва да бъде по-голямо от 1 върху епсилон.
Като по същество сега го доказахме.

Portuguese: 
de menos 1 elevado à n mais 1 sobre n
tem que ser menor que épsilon.
Este menos 1 elevado à n mais um,
este numerador apenas alterna entre valores
negativos e positivos e 1 sobre n.
Mas se tomar o valor absoluto dele,
isto sempre será positivo.
Então isto será o mesmo que 1 sobre n,
cujo valor absoluto tem que ser menor que épsilon.
Agora, n sempre será positivo,
n começa em 1 e vai ao infinito.
Então este valor será sempre positivo.
Isto é dizer que 1 sobre n tem que ser
menos que épsilon para que
isto seja verdade
E agora podemos tirar a recíproca dos dois lados.
Ao tirar a recíproca dos lados de uma desigualdade
teremos que n -- ao tirar a recíproca
dos dois lados de uma desigualdade
temos que inverter o sinal --
Então para isto ser verdade, n tem que
ser maior do que 1 sobre épsilon.
E basicamente concluímos a prova.

English: 
So now we've said, look, for
this particular sequence,
you give me any
epsilon, and I'm going
to set M to be 1 over epsilon.
Because if n is greater than
M, which is 1 over epsilon,
then we know that this right
over here is going to be true.
That is going to be true.
So the limit does
definitely exist.
And so over here, for
this particular epsilon,
it looks like we've picked
0.5 or 1/2 as our epsilon.
So as long as n is greater
than 1 over 1/2, which is 2,
so in this case we could
say, look, you gave me 1/2.
My M is going to be a
function of epsilon.
It's going to be defined
for any epsilon you give me
greater than 0.
So here, 1 over 1/2
is right over here.
I'm going to make my
M right over here.

Thai: 
ตอนนี้เราบอกว่า ดูสิ สำหรับลำดับนี้
คุณให้เอปซิลอนใดๆ มา ผมจะ
ให้ M เท่ากับส่วนเอปซิอลน
เพราะถ้า n มากกว่า M 
ซึ่งก็คือ 1 ส่วนเอปซิลอน
แล้วเราจะรู้ว่า ประโยคนี้ต้องเป็นจริง
มันจะเป็นจริง
ลิมิตมีอยู่จริง
แล้วตรงนี้ สำหรับเอปซิลอนเฉพาะค่านี้
มันดูเหมือนว่า เราได้เลือก 0.5 หรือ 1/2
เป็นเอปซิลอนของเรา
ตราบใดที่ n มากกว่า 1 ส่วน 1/2 ซึ่งก็คือ 2
ในกรณีนี้ เราบอกได้ว่า ดูสิ คุณให้ 1/2 ผมมา
M ของผมจะเป็นฟังก์ชันของเอปซิลอน
มันจะกำหนดค่าตามเอปซิลอนใดๆ 
ที่คุณให้ผมมา
มากกว่า 0
ตรงนี้ 1 ส่วน 1/2 คือค่าตรงนี้
ผมจะสร้าง M ตรงนี้

Korean: 
어떤 수열이 있고
아무 ε이나 주어진다면
M=1/ε이 되는 M을 정할겁니다
n>M이면 n>1/ε이니까
|a_n-0|<ε이 참이 되겠죠
따라서 극한값이 존재하게 됩니다
그래프를 다시 보면
이 경우엔 ε이 1/2쯤 돼보이네요
n>1/(1/2), 
즉 n>2일 때를 볼까요
ε으로 1/2이 주어졌고
M=1/ε이라고 정했습니다
0보다 큰 모든 ε에 대해
성립하죠
1/(1/2)는 여기 쯤이네요
M을 여기에 표시합니다

Portuguese: 
Agora dissemos que, para esta sequência em particular
dado qualquer épsilon,
eu vou estabelecer M como 1 sobre épsilon.
Porque se n for maior que M,
M é o mesmo que 1 sobre épsilon,
saberemos que isto aqui é verdade.
Que isto será verdade.
Então o limite com certeza existe.
Então aqui, para este épsilon em particular,
parece que escolhemos 0,5 como nosso épsilon.
Desde que nosso n seja maior que 1 sobre 1/2,
que é 2, digamos que você me dê 1/2.
Meu M será uma função de épsilon.
Será definido por qualquer épsilon
maior que 0.
Então aqui 1 sobre 1/2 está logo aqui.
Vou fazer o meu M bem aqui.

Bulgarian: 
Казахме, че за тази определена редица,
ако ми дадеш някакво епсилон, аз
ще намеря М, което да е 1 върху епсилон.
Защото ако n е по-голямо от М, което е 1 върху епсилон,
ще знаем, че това тук ще бъде вярно.
Това ще бъде вярно.
Така че границата определено съществува.
Така че тук, за това определено епсилон
изглежда че сме избрали 0,5 или 1/2 за нашето епсилон.
Стига n да е по-голямо от 1 върху 1/2, което е 2,
така че в този случай можем да кажем, че ако ми дадеш 1/2,
М ще бъде функция от епсилон.
Това ще бъде определено за всяко епсилон, което е
по-голямо от 0.
1 върху 1/2 е точно ето тук.
Ще направя М да е ето тук.

Bulgarian: 
Виждаш, че това наистина е случай,
в който редицата е в рамките на обхвата,
когато минава през всяко n по-голямо от 2.
За n равно на 3 тя се намира в обхвата.
За n равно на 4 тя е в обхвата.
За n равно на 5 -- като тя продължава нататък.
 
Ние го доказахме ето тук.
Извършихме доказателството.
Ако ми дадеш всяко друго епсилон,
казах че М е равно на 1 върху това нещо.
За всяко n по-голямо от това, това ще бъде вярно.
Следователно това определено е така.
Тази редица е сходяща към 0.

Thai: 
และคุณเห็นว่ามันเป็นจริง
ที่ลำดับผมอยู่ในช่วง
เมื่อเราผ่าน n มากกว่า 2 ไป
สำหรับ n เท่ากับ 3 มันอยู่ในช่วง
สำหรับ n เท่ากับ 4 มันก็อยู่ในช่วง
สำหรับ n เท่ากับ 5 -- มันยังอยู่ต่อไปเรื่อยๆ
และเราไม่ได้พูดลอยๆ อีกต่อไป
เราได้พิสูจน์แล้วตรงนี้
เราพิสูจน์แล้ว
คุณให้เอปซิลอนอื่นมา
ผมก็บอกว่า M เท่ากับ 1 ส่วนค่านั้น
แล้วสำหรับ n ที่มากกว่านั้น ประโยคนี้เป็นจริง
ประโยคนี้เป็นจริงแน่นอน
ลำดับนี้ลู่เข้าหา 0

English: 
And you see it is
indeed the case
that my sequence
is within the range
as we passed for any
n greater than 2.
So for n is equal to
3 it's in the range.
For n is equal to 4
it's in the range.
For n equals 5-- and it
keeps going and going.
And we're not just
taking our word for it.
We've proven it right over here.
So we've made the proof.
You give me any
other any epsilon,
I said M is equal to
1 over that thing.
And so for n greater than
that, this is going to be true.
So this is definitely the case.
This sequence converges to 0.

Portuguese: 
E pode ver que é o caso
que minha sequência está nesta região
conforme passamos para um n maior que 2.
Isto será verdade para n igual a 3,
e também para n igual a 4.
E também para n igual a 5-- e continua.
Não estamos apenas acreditando cegamente.
Já provamos isto bem aqui.
Fizemos a prova.
Podemos ter qualquer outro épsilon,
eu disse que M é igual a 1 sobre aquilo.
Então para qualquer n maior que aquilo, isto será verdade.
Este é o caso.
Esta sequência converge a 0.
[legendado por: Guilherme Hubner
revisado por: Musa Morena Marcusso Manhães]

Korean: 
이제 확인할 수 있네요
n>2일 때, 이 수열은
색칠된 영역에 포함됩니다
n=3일 때도 영역 안에 있고
n=4여도 영역 안이고
n=5일 때 등등
마찬가집니다
그냥 그렇겠거니 하는게 아닙니다
이쪽에 이미 증명했죠
근거 있는 이야기입니다
어떤 ε이 주어져도
M=1/ε이고
n>M이면
|a_n-0|<ε은 참입니다
이 수열이 그에 해당하죠
이 수열은 0으로 수렴합니다
