# Analysis: Least Upper Bound

1. May 15, 2014

### teme92

1. The problem statement, all variables and given/known data

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

3. 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.

2. May 15, 2014

### pasmith

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

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

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

Let $$y = \frac{14n + 9}{16n + 13}$$ 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 $n$.

3. May 15, 2014

### 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?

Last edited: May 15, 2014
4. May 15, 2014

### Ray Vickson

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

5. May 16, 2014

### teme92

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

6. May 16, 2014

### pasmith

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

7. May 16, 2014

### HallsofIvy

Staff Emeritus
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?

8. May 16, 2014

### 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?

Last edited: May 16, 2014
9. May 16, 2014

### HallsofIvy

Staff Emeritus
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 $\frac{14n+ 9}{16n+ 13}> \frac{7}{8}$?

10. May 16, 2014

### 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?

11. May 16, 2014

### HallsofIvy

Staff Emeritus
Yes, trying to solve $\frac{14n+ 9}{16n+ 13}> \frac{7}{8}$ 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. May 16, 2014

### Ray Vickson

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