Question

determine if the infinite sequence is bounded. a sub n =n/(2^{n+2})

Answer 1

\[A_{n}=\frac{ n }{ 2^{n+2} }\]

Answer 2

\[\frac{ n }{ 2^{n+2} } \le \frac{ n }{ 2^{n+1} }\]

Answer 3

\[n(2^n +2) \le n(2^n +4)\]

Answer 4

\[2n \le 4n\]

Answer 5

\[2 \le 4\]

Answer 6

This is a true statement and it proves that the sequence is monotonic because it is a non-increasing sequence. Furthermore
\[A_{1}=\frac{1}{8}\]

Answer 7

which means it is bounded above by \[\frac{1}{8}\]

Answer 8

how do you show the lower limit? it has to be algebraically, I can not just plug in numbers to the sequence to show that it approaches zero, I have to use algebraic justification with the argument

Answer 9

@TuringTest please tell me you know how to do these, I have been trying to get help with these all day, no one remembers them

Answer 10

gosh i remember doing stuff like this. there is a test where you can tell if the function is converging by proving it algebraically, but i forgot the name of the test :(

Answer 11

maybe try the ratio test to prove it?

Answer 12

no It kind of looks like the squeeze theorem that is used but I can't manage to pull it off on this one because its lower bound is one eight and the upper bound can't be the opposite since this is a sequence the integers are the only numbers used for n so it can't just star off between negative one eight and positive one eighth

Answer 13

@satellite73 please help!!!!

Answer 14

I have been on here all day trying to get help on this concept

Answer 15

i am not sure what you are asking

Answer 16

you have that it is monotone decreasing right?

Answer 17

There is a theorem that says if a sequence is bound and monotonic then it converges. I have shown the proof that it is monotonic

Answer 18

but now I have to show it is bounded

Answer 19

bounded above by the first term, and bounded below by zero

Answer 20

I only have one example to learn it from and when I try to use reasoning like the fact that it is a fraction with an exponential term in the denominator, and the n has to be an integer proves that it can't be lower than zero, but that gets counted as wrong by my professor because I have to prove it algebraically. The solution to this one is that it is monotonic and the proof of that is the same as mine yet the sequence is bounded and they say the absolute value of a sub n is less than one eighth

Answer 21

If you can take a minute I can show you an example we were given as to how to prove what they are bounded by.

Answer 22

determine if \[A_{n}=\frac{ 3n }{ n+2 }\] is monotonic and if it is bounded

Answer 23

i can try but i am honestly lost
it is pretty damn clear that \(n\times 4^{-n}\) is monotone decreasing and
\[\lim_{n\to \infty}n\times 4^{-n}=0\]

Answer 24

\[A_{n} \le A_{n+1}\]

Answer 25

I agree with you @satellite73 but my professor is very strict in what she requires

Answer 26

\[\frac{ 3n }{ n+2 } \le \frac{ 3(n+1) }{ (n+1)+2 }\]

Answer 27

then maybe you have to work from the definition:

\[\lim_{n\to \infty}a_n=0\iff \forall \epsilon >0 \exists N \ni \forall n>N |a_n|<\epsilon\]

Answer 28

in the case above you are trying to prove i assume that the limit is 3
here you are trying to prove it is zero
idea is probably the same

Answer 29

when you cross multiply and combine like terms you end up with \[0 \le 6\] and this proves it is monotonic because it gives a true statement. We are allowed only to plug in n=0, or n=1 depending on the situation for one limit. so in this case \[A_{1}=1 \] and \[1 \le a_{n}\]

Answer 30

no we never did them with epsilon delta proofs and we were told to skip that section.

Answer 31

oh crap
well then how did you prove that the limit was 3 ?

Answer 32

you know that it is monotone decreasing, that the terms are positive and bounded above by \(\frac{1}{8}\) in this case
therefore you know that it has a limit \(L\) of some kind, and \(L\geq0\)
to prove that the limit is in fact zero requires something for sure

Answer 33

if a sequence is monotonic it is either non-increasing or non-decreasing. If it is non-increasing and bounded then the lower bound is the limit, if it is non-decreasing and bounded the limit is the upper bound, and that is why I have to find the least upper bound or least lower bound.

Answer 34

To finish the example I was given,she said we have shown one bound and all that is left is to prove that:
\[A_{n} \le N\] for all n.

Answer 35

right, although you mean "greatest lower bound"
one lower bound is 0
that is the greatest lower bound, because for example another lower bound is -2

Answer 36

\[n \le n+2\] and \[\frac{ n }{ n+2 } \le 1\]

Answer 37

therefore \[\frac{ 3n }{ n+2 } \le 3\] so \[1 \le \frac{ 3n }{ n+2 } \le 3\]

Answer 38

in this example your sequence is increasing to 3

Answer 39

I just say the least lower and least upper bound because that is exactly what the professor said. I asked the same question and she said that it was called least, no explanation why just moved on

Answer 40

the fact that the limit is 3 is completely obvious, on the other hand just because it is bounded above by 3 doesn't mean the limit is 3
it could be for example 2.5

Answer 41

"greatest lower bound"
"least upper bound"
that is what you should know

Answer 42

well the idea as I understand it is that you could pick a million and one functions or numbers that are greater than the nth term of the sequence but when you do it this way you are supposed to pick the one that perfectly bounds the sequence. and if the sequence is non increasing then it cannot oscillate and would therefore have a limit at the greatest lower bound

Answer 43

the same would be true of a non decreasing sequence except the limit would be the least upper bound

Answer 44

the greatest lower bound L means this
L is a lower bound, i.e. \(\forall n\in Z+,a_n\geq L\) and
if M is any other lower bound, then \(L\geq M\)

Answer 45

that is why it is called the GREATEST lower bound , not the "least lower bound"

Answer 46

in your case, the one for the question posted, not the example, the greatest lower bound is zero

Answer 47

and you would prove this by showing that any positive number is NOT a lower bound, i.e. that for all \(\epsilon>0\) you can find an \(n\) for which \(a_n<\epsilon\)

Answer 48

I don't understand that. Our professor took four weeks off during a semester because a family member of hers passed away. No one took the class over and we were all given in completes and told to complete the calculus two class independent study so quite a bit of this is confusing, and she was not very receptive to questions like that so I just accept what she says and look for the answers on my own. Her student assistant is worse than her and gives you a "you haven't studied or done any research" as a responce to 90% of the questions, so I am not callenging you in anyway, you have helped me quite a bit before and I am just trying to figure this all out.

Answer 49

I though that zero was the lower bound as well.

Answer 50

ok not in the example, but in your questions, you have shown that the sequence is monotone decreasing

since the terms are all positive that means for sure that zero is a lower bound
but that in itself does not mean that 0 is the "greatest lower bound" or the limit (although it is) it could be for example \(0.1\)

Answer 51

The textbook shows it as -1/8 and I even tried a calculator to check and at x=0 the f?(x)=0. An email to the student assistant questioning it got this: " I do all the problems and if I find any issues I send a mass email out to all the students, so the problem you are talking about is correct. You are probably just burned out, take a break and try again."

Answer 52

in the example you gave you have the inequality \[1 \le \frac{ 3n }{ n+2 } < 3\] and that the sequence is monotone increasing
but that in itself does not guarantee that the limit is 3, only that the limit is less than or equal to 3

Answer 53

now you have lost me
\[n\times 4^{-n}\geq 0\]

Answer 54

where did that come from?

Answer 55

sorry that is a mistake, it is \[\frac{n}{4\times 2^n}\] or
\[2^{-2-n}n\]

Answer 56

in any case it is positive, and never \(-\frac{1}{8}\)

Answer 57

unless for some reason \(n\) can be negative

Answer 58

I agree i thought it was a mistake but the student assistant said it was not, but he woud not explain why.

Answer 59

And as I understand it n a sequence is from one, or zero to infinity so it could not be negative

Answer 60

are you 100% sure you have the question written correctly ?
\[a_n=\frac{n}{2^{n+2}}\]?

Answer 61

i wouldn't beat my head against the wall for this one
you showed it was decreasing
you then know that the least upper bound is the first term, namely \(\frac{1}{8}\) and you know it is bounded below by zero
therefore it has to have a limit
that the limit is in fact zero requires some small amount of work, but since you are not using epsilons i am not sure how you can do it correctly

Answer 62

i guess i should say "monotone decreasing after the first two terms" as they are both \(\frac{1}{8}\)

Answer 63

I don't know either, thanks for your help