Is the Sequence a_n = n / (2^(n+2)) Monotonic and Bounded?

[physics=world]

1. Determine whether the sequence with the given nth term is monotonic & bounded.

a_n = (n) / (2^(n+2))2.
b_n < b_n+1

3.

(n) / (2^(n+2)) < (n+1) / (2^(n+3))

I multiply both side by (2^(n+2)) and (2^(n+3))
(n)(2^(n+3)) < (n+1)(2^(n+2))
Then i distribute and got:

(n)(2^(n+3)) < (n)(2^(n+2)) + (2^(n+2))
this is where I am stuck. where do i go from here?

 

[Mark44]

1. Determine whether the sequence with the given nth term is monotonic & bounded.

a_n = (n) / (2^(n+2))





2.
b_n < b_n+1




3.

(n) / (2^(n+2)) < (n+1) / (2^(n+3))

You can't start with the inequality above, since that's what you want to end with if the sequence is monotonically decreasing. You can use the fact that 2^{n + 3} = 2 * 2^{n + 2}.

I multiply both side by (2^(n+2)) and (2^(n+3))



(n)(2^(n+3)) < (n+1)(2^(n+2))



Then i distribute and got:

(n)(2^(n+3)) < (n)(2^(n+2)) + (2^(n+2))



this is where I am stuck. where do i go from here?

 

[physics=world]

You can't start with the inequality above, since that's what you want to end with if the sequence is monotonically decreasing. You can use the fact that 2^{n + 3} = 2 * 2^{n + 2}.

so just by knowing that 2^{n + 3} < 2 * 2^{n + 2} is good enough?

 

[Mark44]

so just by knowing that 2^{n + 3} < 2 * 2^{n + 2} is good enough?

No, and it didn't say that 2^{n + 3} < 2 * 2^{n + 2}.

2^{n + 3} = 2 * 2^{n + 2}

 

[physics=world]

okay. so what's next or is that it?

 

[LCKurtz]

1. Determine whether the sequence with the given nth term is monotonic & bounded.

a_n = (n) / (2^(n+2))

Below you are starting out assuming $a_n < a_{n+1}$ which, as Mark pointed out, you don't know. Maybe you could try $a_n\, ?\, a_{n+1}$ and leave the inequality unknown to explore and see where it leads.

(n) / (2^(n+2)) < (n+1) / (2^(n+3))

I multiply both side by (2^(n+2)) and (2^(n+3))

(n)(2^(n+3)) < (n+1)(2^(n+2))

At that step why don't you try dividing out all the 2's you can. Maybe you can get to where you can decide if either ? = < or ? = > makes a true statement. Then what you need to do is start with that true statement and see if you can work backwards to what you are trying to prove.

 

[physics=world]

At that step why don't you try dividing out all the 2's you can. Maybe you can get to where you can decide if either ? = < or ? = > makes a true statement. Then what you need to do is start with that true statement and see if you can work backwards to what you are trying to prove.

okay . so if i divide out the 2's then i get

((2^n+3) / (2^n+2)) ? (n+1) / (n)

2 ? ((n+1) / n)

at this point, I am i doing it right

 

[LCKurtz]

At that step why don't you try dividing out all the 2's you can. Maybe you can get to where you can decide if either ? = < or ? = > makes a true statement. Then what you need to do is start with that true statement and see if you can work backwards to what you are trying to prove.

okay . so if i divide out the 2's then i get

((2^n+3) / (2^n+2)) ? (n+1) / (n)

2 ? ((n+1) / n)

at this point, I am i doing it right

You tell me. Have you simplified it enough that you know whether to use < or > ? What if you multiply both sides by n and keep simplifying?

 

[physics=world]

i would get n ? 1

 

[LCKurtz]

You tell me. Have you simplified it enough that you know whether to use < or > ? What if you multiply both sides by n and keep simplifying?

i would get n ? 1

What about my other question?

 

[physics=world]

yes i have simplify enough.

but how i do i know if its < or >?

 

[LCKurtz]

OK. Now do you see how to start with "if $n>1$ then" and end with the statement that your sequence is monotone and whether it is increasing or decreasing?

 

[LCKurtz]

yes i have simplify enough.

but how i do i know if its < or >?

Which one is true in your problem? n > 1 or n < 1? You need to start with a true statement and work backwards now.

 

[physics=world]

would i just plug in values for n which is n>1 and compare the values to see it its monotonic?

 

[physics=world]

0.125 > 0.093 > 0.0625 > 0.039 ...

when i plug in values n > 1 into the function

 

[LCKurtz]

would i just plug in values for n which is n>1 and compare the values to see it its monotonic?

0.125 > 0.093 > 0.0625 > 0.039 ...

when i plug in values n > 1 into the function

I'm out of time and you obviously aren't getting the point of this discussion. Maybe Mark44 will take it from here. I have to go now.

 

[Mark44]

physics=world,
Now that you have played with this for a while, what's your sense here?
Is a_{n + 1} > a_n?
Or is a_{n + 1} < a_n?

 

[physics=world]

an + 1 < an

 

[Mark44]

OK, so now let's see if we can prove it.

I'm going to take a different tack than LCKurtz did, since you didn't seem to be following what he was doing.

You would like to show that a_{n + 1} < a_n

Here's the start of the argument.
$$ a_{n + 1} = \frac{n + 1}{2^{n+3}} =(1/2)\frac{n + 1}{2^{n + 2}} = (1/2)[\frac{n}{2^{n+2}} + \frac{1}{2^{n+2}}] $$

We would eventually like to come out of this with "< a_n". What can you do with the right-most part of the equation above?

 

[physics=world]

OK, so now let's see if we can prove it.

I'm going to take a different tack than LCKurtz did, since you didn't seem to be following what he was doing.

You would like to show that a_{n + 1} < a_n

Here's the start of the argument.
$$ a_{n + 1} = \frac{n + 1}{2^{n+3}} =(1/2)\frac{n + 1}{2^{n + 2}} = (1/2)[\frac{n}{2^{n+2}} + \frac{1}{2^{n+2}}] $$

We would eventually like to come out of this with "< a_n". What can you do with the right-most part of the equation above?

can we simplify?

 

[Mark44]

can we simplify?

That's not a good idea.