Prove that a sequence is increasing: u(n) = (4n^2 + 1) / (n^2 + 1)

[Qwertyuiop[]]

Hi, I have a sequence u(n) = $\frac{4n^2+1}{\:n^2+1}$. And i have to show that this sequence is increasing. I started by writing it in this form: $4-\frac{3}{n^2+1}$ which was the first part of the question. To show if a sequence is increasing, i know that un+1 - un >0. After simplifying the expression get :$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}\:=\:3\left(\frac{1}{n^2+1}-\frac{1}{\left(n+1\right)^2+1}\right)$. I am not sure how to prove that it is positive.

 

[topsquark]

...$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}$.

The second fraction has a bigger denominator (and the same numerator) as the first. So the second fraction is smaller.

You could formalize this by setting
$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1} = x$

$3((n+1)^2 + 1) - 3(n^2 + 1) = x (n^2+1)((n+1)^2+1)$

$6n = x (n^2+1)((n+1)^2+1)$

Now, both of 6n and $(n^2+1)((n+1)^2+1)$ are positive, so therefore so is x.

Or you could be more formal still, and do an induction proof, using this step.

-Dan

 

[lev888]

Hi, I have a sequence u(n) = $\frac{4n^2+1}{\:n^2+1}$. And i have to show that this sequence is increasing. I started by writing it in this form: $4-\frac{3}{n^2+1}$ which was the first part of the question. To show if a sequence is increasing, i know that un+1 - un >0. After simplifying the expression get :$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}\:=\:3\left(\frac{1}{n^2+1}-\frac{1}{\left(n+1\right)^2+1}\right)$. I am not sure how to prove that it is positive.

Consider $4-\frac{3}{n^2+1}$. As n increases what happens to the denominator? What happens to the fraction? And to the whole expression?
As for the difference of 2 consecutive terms - you have a difference of 2 fractions. Can you make the denominators the same and subtract the fractions? The result should be positive. (you need to switch the order of fractions in your expression)

 

[Steven G]

If u_n/u_n+1> 1, then what is true?

Using your method I would have proceeded as follows:

If$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}\:>0$

Then, $\frac{3}{n^2+1}>\frac{3}{\left(n+1\right)^2+1}\:$

Then $\displaystyle \dfrac{(n+1)^2+1}{n^2+1}>0$
Continuing from here should not be too hard.

 

[Dr.Peterson]

If$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}\:>0$

Then, $\frac{3}{n^2+1}>\frac{3}{\left(n+1\right)^2+1}\:$

Then $\displaystyle \dfrac{(n+1)^2+1}{n^2+1}>0$
Continuing from here should not be too hard.

Don't you mean $\displaystyle \dfrac{(n+1)^2+1}{n^2+1}>{\color{Red}1}$? Presumably you multiplied both sides by the denominator on the right, and divided by 3; this is legal because both are positive.

For an inequality of this sort where some factors are not necessarily positive, it is preferred to leave 0 on the right and combine fractions, I would stick with that method, which isn't very hard.

 

[Qwertyuiop[]]

If u_n/u_n+1> 1, then what is true?

Using your method I would have proceeded as follows:

If$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}\:>0$

Then, $\frac{3}{n^2+1}>\frac{3}{\left(n+1\right)^2+1}\:$

Then $\displaystyle \dfrac{(n+1)^2+1}{n^2+1}>0$
Continuing from here should not be too hard.

Yes, in this form , It's clear that the fraction is positive because (n+1)^2 >=0 and n^2+1 is strictly positive so the expression on the left must be greater than 0.

 

[Dr.Peterson]

Yes, in this form , It's clear that the fraction is positive because (n+1)^2 >=0 and n^2+1 is strictly positive so the expression on the left must be greater than 0.

Please check the details of what he said, and don't trust it until you've done the work yourself, and get the same result. (You won't.)

I'd say the same about my own advice ... which is different from his.

 

[Steven G]

Don't you mean $\displaystyle \dfrac{(n+1)^2+1}{n^2+1}>{\color{Red}1}$? Presumably you multiplied both sides by the denominator on the right, and divided by 3; this is legal because both are positive.

For an inequality of this sort where some factors are not necessarily positive, it is preferred to leave 0 on the right and combine fractions, I would stick with that method, which isn't very hard.

I tend to agree with you but in this case it is just so obvious that everything is positive.

 

[Dr.Peterson]

I tend to agree with you but in this case it is just so obvious that everything is positive.

But my point is that your inequality is wrong. It's not just a matter of being positive! Please look again at your work:

If$\frac{3}{n^2+1}-\frac{3}{\left(n+1\right)^2+1}\:>0$

Then, $\frac{3}{n^2+1}>\frac{3}{\left(n+1\right)^2+1}\:$

Then $\displaystyle \dfrac{(n+1)^2+1}{n^2+1}>0$

The last line is incorrect.

Now, the problem is almost trivial in almost every way you look at it, including the corrected version of this way; but that doesn't mean you can do it a wrong way.

 

[Steven G]

But my point is that your inequality is wrong. It's not just a matter of being positive! Please look again at your work:


The last line is incorrect.

Now, the problem is almost trivial in almost every way you look at it, including the corrected version of this way; but that doesn't mean you can do it a wrong way.

Yes, you have already pointed out that the last line is wrong. I understand that it should be 1