Question

How do you know if a sequence is bounded( lower or upper bounded) ? Ex. An = n/(2^(n+2))

Answer 1

Take the limit as x approaches infinity, nay?

Answer 2

yes that is the right way

Answer 3

f the limit exist the sequence is bounded

Answer 4

Let me revise that statement I just made: If the sequence converges, it's bounded. If it diverges, it doesn't tell you anything.

Answer 5

?
Can't you say, if it diverges, that it's unbounded?

Answer 6

no each convergence sequence is bounded and vice versa

Answer 7

yes

Answer 8

Oh wait wait. It could diverge but still be bounded.

Answer 9

Take the sequence \[\sum_{n=0}^\infty ({-1})^n\]This is bounded but diverges.

Answer 10

Wait, my book says that if a sequence is bounded and monotonic, it is convergent. My question is how to figure out if its bounded or not to find if its convergent?

Answer 11

Right. Good example, Georgey.

Answer 12

i think you are not right

Answer 13

If it's monotonic and bounded, it is definitely convergent. My example is clearly not monotonic, which is why it's divergent.

Answer 14

To prove it's bounded, you would need to find some thing (anything) that is greater than every term of the sequence.

Answer 15

Or it could be bounded on the lower end, in which case just find something that's less than every term in the sequence.

Answer 16

So...how do you find out where it is bounded at?

Answer 17

For this sequence, notice that it's monotonically decreasing. Does it every reach 0?

Answer 18

Pick 0 as a lower bound, and it's pretty easy to prove that any arbitrary term of the sequence is greater than 0.

Answer 19

find the limit n tends to \[\infty\]

Answer 20

If it's monotonic, just take the first term. It's either the lower bound or the upper bound depending on if the sequence is monotonically increasing or decreasing.

Answer 21

Unless the first term is \(\pm \infty\). In that case, take the limit as \(n \rightarrow \infty\)

Answer 22

Eh... the first term doesn't do it for you...
if it's monotonically increasing, you need an upper bound to say that the limit exists.
If it's monotonically decreasing, you need a lower bound to say that the limit exists.
The first term will be a bound, but not the bound you need.

Answer 23

hm..could you explain how i would find the bound for...say... (-1)^n(1/n)

Answer 24

i think it is convergent to 0

Answer 25

I think the best answer, in my experience is just to take some terms of the sequence and see if you can notice the pattern, what is happening with it.
With the example you gave, I get
A1 = 1/8
A2 = 2/16 =1/8
A3 = 3/32
A4 = 4/64 = 1/16
A5 = 5/128
A6 = 6/256
A7 = 7/512

As the denominator grows exponentially, the terms are getting very very small quickly. So I can see that the sequence decreases monotonically. If I can say that there's a lower bound that it never passes, then you can say that the limit exists. If you can say specifically what the GREATEST lower bound is, then THAT will be the limit.

So, I notice it's getting quite small, but will never be negative, so the lower bound to pick becomes obvious. Pick 0, and prove that it's a lower bound.

Answer 26

so as i said befor you find it convergent to 0

Answer 27

For (-1)^n(1/n)
A1 = -1
A2 = 1/2
A3 = -1/3
A4 = 1/4
A5 = -1/5
A6 = 1/6
and so on.