#sequence problem
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you want to know why the options shown are correct/incorrect?
No no. I want to understand all the options
And the topic name related to it
I want to learn the method actually
oh you dont know what they mean?
I only know about convergent and divergent
well you can look up subsequence, bounded, and cauchy sequence
what don't you know
What is wrong in option A
if there is just one convergent subsequence that does not mean the initial sequence is bounded
e.g 1 2 1 3 1 4 1 5 ...
for ease of reading, write out the problem formulation in your question or take better pictures of it
If you download the picture then it will be more clear if we have problems with pictures
why are you doing analysis without studying sequences
what
I have studied sequences and solved some of their examples/questions
did you read the books I've recommended
does it not cover cauchy sequences
Books no
or subsequences
why?
then this question really has no place for confusion
Notes me padha hu bhai cauchy k nare me
do you know what a subsequence is
well let's look at 1
"every sequence that has a convergent subsequence is bounded"
if we had "sequence" instead of "subsequence" it'd be true
(prove it)
but as it's a subsequence! the rest of the sequence can go fuck itself
Coffeλ
cos(1/n) ke baare me kia bol sakte?
cos(x) 1 se 0 ke beech me kaisa behave karta?
,w graph cos(x)
Oscillation karega?
increasing strictly between 1 and zero (zero to 1)
ulta bola
1 to zero
toh cos(1/n), 1 ke aas paas ayega
but n toh bas baare jaa raha he na?
is liye sequence hi monotone increasing he.
4 thoda tricky he
but it's exactly similar to 1
it has a cauchy subsequence i.e. the convergent subsequence!
because convergence => cauchy :3
but the whole sequence might not be cauchy
this example suffices
Bhai ye example kha se aaya apko dimag me?
aise hi aagya

ek sequence ke baare me socha jisme se ek part, ek index baad karke zero hojata
@exotic sparrow aur bhi ek accha example de raha
First we look at the set
$\qty{\frac{1}{m}+\frac{1}{n} : m, n \in \mathbb{N}_{>0}}$
Coffeλ
this is countable
we want to make a sequence that covers every term here
how would we do it
what's your experience with math @exotic sparrow your questions are so often about definitions and basic technique. are you an undergrad? high schooler?
we can do this via
$a_{11},a_{1,2},a_{2,1},a_{1,3},a_{2,2},a_{3,1},\ldots$
@iron tide do you know of a simple closed form expression for each term
Coffeλ
it's just the usual way to sequencize! the Cartesian product of 2 countable sets
$\qty{\frac{1}{m}+\frac{1}{n} : m, n \in \mathbb{N}_{>0}}$ is just like $\mathbb{N}\times \mathbb{N}$
without the zeroes
how do i say this without butchering notation
you mean like go over diagonals?
Coffeλ
yes
$1/m+1/n$ can be compared to $(m, n)$
Coffeλ
and guess what if we use the set $\qty{n+1/m : n, m \in \mathbb{N}_{>0}}$ and put it in a sequence, we get a diverging sequence with infinite converging subsequences!
Coffeλ
$\qty{n+1/m:n,m\in \mathbb{N} _{>0}} = \bigcup {n\in \mathbb{N}\setminus \qty{0}}\qty{n+1/m : m\in \mathbb{N}{>0}}$
@exotic sparrow you there?
Yes. I am here
oh no
Coffeλ
so a sequence can have a lot of converging subsequences
$$ \sum _{i=0}^\infty \sum _{j=0}^\infty a(i,j) = \sum _{i=0}^\infty \sum _{j=0}^i a(i-j,j) $$
aL
if that's what you meant
I want a closed expression for each term
this is using 2 nested sequences isn't it
well ya it's counting elements of Q along the diagonals basically
i get that, can we put it into 1 sequence
so I can show it to Arjun
i guess
it'll be a countable union of pieces of length 1, 2 , 3 etc
mhm
anyways
I'm done lazing around for now
@exotic sparrow you done?
as in do you understand now
I am still here
samajh me aaya?
Ha aa toh raha hain lekin ek bar me toh nahin mujhe time chahiye aur bar bar pdna pdega

wdym
he said "i can understand it but not in one reading, ill have to read it over and over over some time"
@exotic sparrow why tho?
For better understanding 😆
what is better understanding to you
I have a slow processor chip
fair enough, as long as you put effort into understanding solutions it's fine
wat
I'm asking you
what is "better understanding" to you
Matlab???
otherwise you will not learn and keep posting questions in the style of "how do I solve this" instead of "I have tried this this and that but it's not working"
Jab tak me us cheez ko realise visuals aur feel na kar leta hoon
"aur bhi acche samajhna" iska matlab kia he tumhare naate
Areee nooo
I didn't say that
You took it wrong interpretation
remember that everything starts at definitions
I said i will read ot twice and thrice so that i will understand it better
learn the definitions and relevant examples
English is more important for you right now
please learn English well otherwise you won't be able to read texts.
Texts toh samaj aa rha hain par me express sahi nahin kar paata hoon
yeah learn English
Really sorry for it

watch movies or something
with captions
will make you English master in a month
I have learnt english so far through reading and talking with people on discord. I am learning a lot by discord. I am graduated in bachelor of science mathematics from government University where professor doesn't come a single day. I studied and got minimum marks so i have a degree but i haven't read proper text books and anything. But now i started again on my own.
@iron tide
how the fuck is you a bsc bro wtf 
Just went and passed papers
In our university exams they just ask previous questions and via rote learning i git the degree 😭

this is reassuring
I can get a degree in math too
yeee
(hopefully)
you have no idea
about the 14 yr Olds
they are crazy they will take over the world
0 chance of my ass getting through
17 in May
it's really not something special, anyone can do it if they want to learn it
ngl thought you were like 25ish
Yes yesss this dude was lying to me
What about you? Al seems PG graduated to me 🫣
A sequence can be made of two and more convergence subsequences
But this is not possible in series i guess
Because the general term will be destroyed m
I'm physically equivalent to one yes
(in a bad way, I shouldn't have spondylitis)
I have an interesting question because it is related to a series which contains these kinds of subsequences
Check
@slender magnet @iron tide
Sr for ping
subsequence?
what's with the wording
but one way or the other
it's related to the ratio test
do you know what $\lim \sup$ is?
Coffe𝑦
@exotic sparrow
Yes. Ratio test
Which shows us it is increasing to a limit or not
When we divide two terms each other then we get a value if greater than one which means like series increasing or diverging so and when we get a value less than 1 inch the series is converging
By mistake I wrote subsequences
So I guess this question is wrong because two different types of sequences in a series is not possible
what.
lim sup isn't ratio test.
wdym.
do you know what the limit supremum is
Ohh the upper bound

then what's sup
$\lim \sup a_n \leq \sup a_n$
Coffe𝑦
more importantly
Bhai pura yehi samjhaate samjhaate time lagjayga
yeh part kitab se karlo
Ruko me likhta hoon
tao wala sahi ho raha? agar woh aasan lage toh fir dusra pagro
Aur supremo ka matlab yah hai ki koi bhi Jaise sequence hai to vah uska jo list upper bound hota hai aur lowest upper bond
Best lag rhi hain ye toh mere level k according
haa ye supremum he
Aur limit laga di sup toh
chalo accha he fir :p
Vo ek number dega
hmm
Agar finite hua toh convergence
1/n
ya choro definition hi dedeta
Coffe𝑦
@exotic sparrow
Coffe𝑦
haa ab Sahi he
ye cheez bohot kaam ki he
ye sequence ke limit point ke bound he
Acha
toh diverging sequence ka sup toh infinity he
but limit supremum na bhi ho sakta he!
Agar limit exists kara toh sequence convergence?
Abhi toh sequence k example dekh rha hu book se 107 page par hoon
me yeh bata raha kiuki mujhe khud agar koi yeh batata toh accha hota
- ek chapter padhke samajhne me time lagega, ye lim sup lim inf mujhe samajh ne me kam se lam 2-3 hafte lage bas 6 page! toh haar mat mano. Agar nahi samajh aye yaha puchna nahi toh time waste hoga.
- har exercise karo waha se aur proof jo exercise me diye he!
- proof writing sikhlo yaha se.
- proof writing haat me laane ke liye proof book se likho khud!
- har din kitna padha dm karke batana mujhe, me extra kuch bata dunga jo tao book me nahi he
@exotic sparrow
hein
volume 2 mat padho abhi
vol 1 ke baad zakon karlo
fir me vol 2 krunga
kiuki vol2 me Fourier series, measure theory wager he
Fourier series ka ratta mar k pass kiya tha 😭

And vo laplase transformation
Mujhe samjh toh sab aa jata haun lekin vhi haina khud se kuch ni kiya kuch
Ungali pakdne ki aadth ho gyi
Bhai ye question me samjh gya
Option B discarded
Cos(1/n)/(1/n) when n tends to infinity
So it will be infinite
what.
I meant option b discarded because iska toh limit infinity ja rha
kiska.
kiska baat kar rahe ho
Vc aa sakte bhai?
Ye question ki
Type kr lio
yeah
okay
konse wale batana specifically
theek he
dikh nahi rha
quality kharab he
ok
nope
mhm
okay
ha
konse proof specifically batana
sin(n)/n ?
are sin(n) ka absolute value 1 se kaam nahi he?
ha
hogya
bounded matlab convergent nahi
bounded monotone matlab convergent
ha
tao se padhe ye?
tao se dekhlo ekbar
woh analysis pe sabse aasan book he
😆 english me padhna practice karo
abhi english level pe lao
kiuki sabkuch ke liye woh chaiye
ok
unke proof dekhe?
$a_n \to p,b_n \to q \implies a_nb_n\to pq$
Coffe𝑦
ye waale
dikhao
haa ekdam
definition bohot important he
ye proof dekha?
bolzano weirstrass ki proof
kisika proof samajhna matlab pura samajhna
me kia batau, tao ke pehle woh sequence khatam karne me 1-1.5 mahine lage
hojayga tension mat lo
JAM?
metric space ha
metric space real analysis me hi he
@sullen turret tu bhi real analsysis kar raha he na?
abhi start kiya?
download krke dekhta
kaha tak kiya?
kya kaam h jldi bol
bata na
proof ghatiye deke rakhe
ye sachme 15 saal ka he.
na
math pursue karni he toh krlete
ye proof
woh waale theorem ka proof
ye left page ka niche wala aur right page ke upar wala
hint deke rakha but yehi he basically proof
$a_m x_m - aq = a_m(x_m-q) + (a_m-a)q$
Coffe𝑦
deke rakhi he x_m -> q aur a_m -> a
ab a_m aur x_m bounded he na?
kiuki convergent he
ha toh aisa ek $M$ he ki $\abs{a_m}, \abs{x_m} <M$
Coffe𝑦
(bounded)
ha dono kuch number se chote he
unme se bara wala lelo
-M < a_m < M
-M < x_m < M
ye aaya
aab inko yaha lagao
bas a_m ke hi lagalo 
haan batao
konsa
ye?
haa ye toh bas rearrange kia jaake x_m - q bhi bohot chota hojaye aur a_m - q bhi
a_m aur q jo bounded he woh unke saath niche kheecha jayga
personally ye cheeze tao me utni acchi nahi he aur proofs kuch baare he
$-M(x_m-q) + (a_m-a)q < s_m < M(x_m-q) + (a_m-a)q$
Coffe𝑦
yaha se squeeze theorem lagalo
s_m ye wala
bounded to he
ye zero pe converge bhi karta
konse
Ye
pehle deke rakha ki x_m q pe converge karta aur a_m a pe
isse jyada confusion hua
2 mahine lagbhag lage aadat aane ki
aur dusre logo ko iske baare me pucha
but abhi jo bata raha ye dusre book se he

