Question

Analysis
Does there exist a sequence
\[{A _{n} : \limsup A _{n}=-\infty} \] ?

Answer 1

an = -n
then this goes to -infinity
so limit = limit sup = lim inf

Answer 2

do you need to prove the limit also?

Answer 3

or the no limit rather....

Answer 4

that's what i thought it might be, but I started confusing myself lol!
Nope no proof, it's just a true/false thing...

It also asks "If a sequence {An} converges to a finite limit point 'a' is {An} bounded?"
I wasn't sure if this was true or false, as I'm not sure if it has to be bounded both above and below in order to be 'bounded'?

Answer 5

every convergent sequence is bounded

Answer 6

but not vice versa

Answer 7

But, every bounded sequence contains a convergent sub-sequence

Answer 8

Ahh yes, that's a theorem, I remember now!

Great, thankyou so much! :)

Answer 9

np good luck:)

Answer 10

Sorry @zzr0ck3r to disturb you again,
for "a convergent sequence with two distinct accumulation points?"
would \[{A _{n}} = (-1)^{n}\frac{ 1 }{ n }\] work?

Answer 11

hmm, so we need the limit at infinity to flip flop?

Answer 12

this is a strange question
if a sequence converges then it converges to some limit L
so it will have only one accumulation point
I thought....

Answer 13

That's what I thought...
but wouldn't the sequence I gave above have two limit points? [ie -1 and 1?]

i may be completely wrong

though this is a true/false question too so it could be false i suppose

Answer 14

I think that would go to 0

Answer 15

The answer is false :)
You were correct, if a sequence converges then it converges to one, and only one, limit point, L.

Thankyou!

Answer 16

it has to go to zero for the series to converge and we know that series converges (alternating harmonic)

Answer 17

unless that theorem is only for A_n > 0

Answer 18

anyway I think it goes to 0 (-1)^n flip flops and does not converge but has limit inf -1 and limit sup 1