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

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Homework Help Overview

The discussion revolves around finding the least upper bound of the set of real numbers defined by the expression E = {14n + 9/(16n + 13): n ∈ N}. Participants explore the behavior of this expression as n varies, particularly focusing on its limits and bounds.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the set E and question whether it has an upper bound. There are attempts to analyze the expression by manipulating it and considering its limit as n approaches infinity. Some participants suggest solving inequalities to determine bounds, while others question the positivity of n in their calculations.

Discussion Status

The discussion is active, with participants providing insights and clarifications about the expression and its limits. There is a recognition that the limit approaches 7/8, and some participants assert that this value serves as the least upper bound, although there is no explicit consensus on the final conclusion yet.

Contextual Notes

There is some confusion regarding the definition of natural numbers (N) and whether they include only positive integers. Additionally, participants are navigating through assumptions about the behavior of the expression as n increases, leading to discussions about convergence and divergence.

teme92
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Homework Statement



Find, with proof, the least upper bound of the set of real numbers E given by:
E ={14n + 9/16n + 13: n \in 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.
 
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teme92 said:

Homework Statement



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


As written you have <br /> E = \{ 14n + \frac{9}{16n} + 13 : n \in \mathbb{N}\}<br /> 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 <br /> E = \{ \frac{14n + 9}{16n + 13} : n \in \mathbb{N} \}.<br />

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 <br /> y = \frac{14n + 9}{16n + 13}<br /> and solve for n. You need n to be positive, so that gives you a bound y_0 on y.

You then need to see whether you can make y_0 - \frac{14n + 9}{16n + 13} arbitrarily small by suitable choice of n.
 
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?
 
Last edited:
teme92 said:
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?

Does N mean all integers (negative and positive) or just the non-negative positive integers?
 
Hey Ray Vickson, N is natural numbers so all positive integers.
 
teme92 said:
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 y &lt; 9/13 and negative when y &gt; 9/13 and the denominator is positive when y &gt; 16/14 and negative when y &lt; 16/14. For the quotient to be positive, the numerator and denominator must have the same sign.
 
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:
\frac{14+ \frac{9}{n}}{16+ \frac{13}{n}}
Now, it is obvious what happens as n goes to infinity. Is y ever larger than that number for finite n?
 
So the limit is 14/16 and as n goes to infinity.

As n increases E decreases so away from L so its divergent?
 
Last edited:
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 \frac{14n+ 9}{16n+ 13}&gt; \frac{7}{8}?
 
  • #10
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?
 
  • #11
Yes, trying to solve \frac{14n+ 9}{16n+ 13}&gt; \frac{7}{8} 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".
 
  • #12
HallsofIvy said:
Yes, trying to solve \frac{14n+ 9}{16n+ 13}&gt; \frac{7}{8} 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
f(x) = \frac{9 + 14x}{13+16x}, \; x \geq 0
 

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