Question

{-n^2} from n=0 to inf , according to text this sequence is bounded above and the upper bound value is 0, now should the upper bound value must be between the given limit that's for this question is n=0 to inf therefore the upper bound value has come to zero if not then the upper bound value could be anything so why not we say it could be -1,-2,-3,-4,...,-inf ?

Answer 1

an upper bound is zero
an upper bound is also 22 or any positive number

Answer 2

how come? -n^2 > -n^2 +1

Answer 3

you wrote
\[\{-n^2\}_{n=0}^{\infty}\] so the seqence should look like
\[\{-1,-4,-9,-16,...\}\]

Answer 4

exactlly

Answer 5

\[\{0,-1,-4,-9,-16,...\}\] sorry that was wrong, it should start at n = 0

Answer 6

ok SO the upper or lower bound value should be from the sequence?

Answer 7

no not necessarily

Answer 8

i just made a mistake, since it says start at n = 0 i need to replace n by 0 first

Answer 9

I mean if the upper and lower limit has been defined then it should be from the sequence? isn't?

Answer 10

so 0 IS an upper bound, but the upper bound is not unique
the least upper bound is uniqe, and that is 0

Answer 11

there is clearly no lower bound for this one, since any number you pick you can always find a number
\[-n^2\] less that it

Answer 12

do we plugin the bound value in sequence to test if it works?

Answer 13

if "a" is an upper bound it should mean
\[a\geq x_i\] for all
\[x_i\in \{x_n\}\]

Answer 14

mostly you find a upper bound by thinking

Answer 15

thinking what? :|

Answer 16

not by a computation, although you may use one
if you have a sequence
\[\{0,-1,-4,-9,-16...\}\] it is pretty clear that 0 is an upper bound, and in fact that 0 is the least upper bound

Answer 17

if you have say
\[\{\frac{1}{n}\}_{n=1}^{\infty}\] you can see that the least upper bound is 1, and the greatest lower bound is 0

Answer 18

do we have to think that if we plugin the bound value in the sequence formula and it satisfies the condition of lower or upper bound values

Answer 19

yes

Answer 20

how did you find the bound value so quickly? is it because it is n=1 to inf?

Answer 21

if you want an upper bound you are trying to find an "a" greater than or equal to all numbers in the sequence

Answer 22

sure just by thinking

Answer 23

{(-1)^n+1} from n 1 to inf the upper bound value is 1 and lower bound value is -1 how did we found these values?

Answer 24

sequence starts
\[1,\frac{1}{2},\frac{1}{3},...\] clearly it goes to zero and clearly 1 is the largest

Answer 25

okkkkkkkkkkkkkkkkkkk

Answer 26

your sequence takes on only two values, 1 and -1

Answer 27

so yes, 1 is the largest and -1 is the smallest

Answer 28

if a equals to every number of the sequence then will it still be the upper bound value?

Answer 29

again it comes from looking at the terms in the sequence and see what you have

Answer 30

think of the silly sequence
\[\{1,1,1,1,...\}\] not a very interesting sequence, a constant sequence.
the upper bound is 1, and likewise the lower bound is 1

Answer 31

so all we have to analyse the sequence values?

Answer 32

why the upper and lower are both same in this silly seq?

Answer 33

because for every element x in the squence both of these statements are true:
\[1\leq x\] and
\[1\geq x\]

Answer 34

ok the theoram says it not only have to be greater or less it could be equal as well

Answer 35

gotcha

Answer 36

yes

Answer 37

thanksssss bro

Answer 38

yw