It has a horizontal asymptote at y = 7/8, and increasing for all x > -13/16.

[teme92]

Homework Statement

Find, with proof, the least upper bound of the set of real numbers E given by:
E ={14n + 9/16n + 13: n [itex]\in[/itex] N}

:

Homework Equations

The Attempt at a Solution

So I said that 16n+13>14n+9 for all N

From this I get n>-2

What do I do with this? I understand that as n increase E will decrease but I don't know how to answer the question. Any help would be much appreciated.

 

[pasmith]

Homework Statement

Find, with proof, the least upper bound of the set of real numbers E given by:
E ={14n + 9/16n + 13: n [itex]\in[/itex] N}


As written you have [tex]
E = \{ 14n + \frac{9}{16n} + 13 : n \in \mathbb{N}\}
[/tex] which has no upper bound.

:

Homework Equations

The Attempt at a Solution

So I said that 16n+13>14n+9 for all N

I infer from this that you actually meant [tex]
E = \{ \frac{14n + 9}{16n + 13} : n \in \mathbb{N} \}.
[/tex]

From this I get n>-2

You should get that 1 is an upper bound for E. But is it the least?

What do I do with this? I understand that as n increase E will decrease but I don't know how to answer the question. Any help would be much appreciated.

Let [tex]
y = \frac{14n + 9}{16n + 13}
[/tex] and solve for [itex]n[/itex]. You need [itex]n[/itex] to be positive, so that gives you a bound [itex]y_0[/itex] on [itex]y[/itex].

You then need to see whether you can make [tex]y_0 - \frac{14n + 9}{16n + 13}[/tex] arbitrarily small by suitable choice of [itex]n[/itex].

 

[teme92]

Hey pasmith,

You're right, that is what I meant (long day).

y = (14n + 9)/(16n + 13)

And you said solve for n so:

n= (9-13y)/(16y-14)

However n would not be postive here?

 

[Ray Vickson]

Hey pasmith,

You're right, that is what I meant (long day).

y = (14n + 9)/(16n + 13)

And you said solve for n so:

n= (9-13y)/(16y-14)

However n would not be postive here?

Does N mean all integers (negative and positive) or just the non-negative positive integers?

 

[teme92]

Hey Ray Vickson, N is natural numbers so all positive integers.

 

[pasmith]

Hey pasmith,

You're right, that is what I meant (long day).

y = (14n + 9)/(16n + 13)

And you said solve for n so:

n= (9-13y)/(16y-14)

However n would not be positive here?

You can see that the numerator is positive when [itex]y < 9/13[/itex] and negative when [itex]y > 9/13[/itex] and the denominator is positive when [itex]y > 16/14[/itex] and negative when [itex]y < 16/14[/itex]. For the quotient to be positive, the numerator and denominator must have the same sign.

 

[HallsofIvy]

I don't see any point in solving for n. You are concerned with values of the fraction, not values of n. Instead, divide both numerator and denominator by n:
[tex]\frac{14+ \frac{9}{n}}{16+ \frac{13}{n}}[/tex]
Now, it is obvious what happens as n goes to infinity. Is y ever larger than that number for finite n?

 

[teme92]

So the limit is 14/16 and as n goes to infinity.

As n increases E decreases so away from L so its divergent?

 

[HallsofIvy]

I am no longer sure what you are talking about. Saying "the limit is 14/16" means the sequence is convergent, doesn't it? It has a limit. Clearly the fraction approaches 14/16= 7/8 as closely as we please. The only question left is "is it ever, for some finite value of n, larger than 7/8:

Can you solve [itex]\frac{14n+ 9}{16n+ 13}> \frac{7}{8}[/itex]?

 

[teme92]

Ok I understand that bit now.

Solving the inequality:

8(14n+9)>7(16n+13)
112n+72>112n+91
72>91

Which doesn't make sense so 7/8 is the least upper bound?

 

[HallsofIvy]

Yes, trying to solve [itex]\frac{14n+ 9}{16n+ 13}> \frac{7}{8}[/itex] leads to a statement that is false for all n (I would not say "doesn't makes sense"- just "false") so the inequality is never true- 7/8 is an upper bound on the fraction. But we also know that it comes arbitrarily close to 7/8 so 7/8 is the "least upper bound".

 

[Ray Vickson]

Yes, trying to solve [itex]\frac{14n+ 9}{16n+ 13}> \frac{7}{8}[/itex] leads to a statement that is false for all n (I would not say "doesn't makes sense"- just "false") so the inequality is never true- 7/8 is an upper bound on the fraction. But we also know that it comes arbitrarily close to 7/8 so 7/8 is the "least upper bound".

Alternatively, use calculus to examine the behavior of the function
[tex] f(x) = \frac{9 + 14x}{13+16x}, \; x \geq 0[/tex]