Proving Sequence a_n = 1 as n Goes to Infinity

[Natasha1]

I can't do this one :-(

Prove from the definition that a_n=\frac{n^2 + 3n}{n^2 +2}<br />-> 1 as n -> \infty

 

[LeonhardEuler]

Try dividing numerator and denominator by n². Then assume \epsilon &gt;0 and find an N so that |a_n-1|&lt;\epsilon whenever n>N.

 

[Natasha1]

Try dividing numerator and denominator by n². Then assume \epsilon &gt;0 and find an N so that |a_n-1|&lt;\epsilon whenever n>N.

My lecturer wrote on the board that we should be answering the question as follows (he never gave us an example though!):

For a given positive h you must ensure that the difference of a_n and 1 can be made to stay less than h. When you have an expression for the difference, look for a related simpler expression that can easily be made less than h.

 

[LeonhardEuler]

Yes, that's the idea. Write out the difference:
|a_n-1|=\frac{n^2 + 3n}{n^2 +2}-1=|\frac{n^2+3n-n^2-2}{n^2+2}|=|\frac{3n-2}{n^2+2}|&lt;h
(your using h instead of \epsilon) Now try to simplify this using some inequalites. I'll show you how to make the first simplification. You want this to be less than h, and you can see that it is less than:
|\frac{3n-2}{n^2}|
so it will suffice to show that this is less than h when n is greater than some N. Since n is positive, you can pull it outside the absolute value to get:
\frac{1}{n}|3-\frac{2}{n}|&lt;h
You just need to make one more simplification and you should be able to get a formula for N interms of h.

 

[Natasha1]

Yes, that's the idea. Write out the difference:
|a_n-1|=\frac{n^2 + 3n}{n^2 +2}-1=|\frac{n^2+3n-n^2-2}{n^2+2}|=|\frac{3n-2}{n^2+2}|&lt;h
(your using h instead of \epsilon) Now try to simplify this using some inequalites. I'll show you how to make the first simplification. You want this to be less than h, and you can see that it is less than:
|\frac{3n-2}{n^2}|
so it will suffice to show that this is less than h when n is greater than some N. Since n is positive, you can pull it outside the absolute value to get:
\frac{1}{n}|3-\frac{2}{n}|&lt;h
You just need to make one more simplification and you should be able to get a formula for N interms of h.

I mean I can do it the way by dividing both numerator and denominator by n^2

Here is my solution:

a_n = (n^2 +3n) / (n^2 +2) --> 1, as n-->infinity.
Divide both numerator and denominator by n^2,
a_n = [(n^2)/(n^2) +(3n)/(n^2) ] / [(n^2)/(n^2) +2/(n^2)] --->1, as n-->infinity.
a_n = [1 +3/n] / [1 +2/(n^2)] --> 1, as n-->infinity.
The 3/n and 2/(n^2) approach zero as n approaches infinity, so,
a_n = [1] / [1] ---> 1
a_n --> 1

But I have never ever done it using that h and N? And I am lost. Would someone mind doing it? To get to a_n -->1 as n --> infinity

 

[LeonhardEuler]

I mean I can do it the way by dividing both numerator and denominator by n^2

Here is my solution:

a_n = (n^2 +3n) / (n^2 +2) --> 1, as n-->infinity.
Divide both numerator and denominator by n^2,
a_n = [(n^2)/(n^2) +(3n)/(n^2) ] / [(n^2)/(n^2) +2/(n^2)] --->1, as n-->infinity.
a_n = [1 +3/n] / [1 +2/(n^2)] --> 1, as n-->infinity.
The 3/n and 2/(n^2) approach zero as n approaches infinity, so,
a_n = [1] / [1] ---> 1
a_n --> 1

But I have never ever done it using that h and N? And I am lost. Would someone mind doing it? To get to a_n -->1 as n --> infinity

You have proven that a_n-->1, but you didn't use the definition. Since you've never seen an example of how its done I'll show ou how to do a similar one. suppose a_n=\frac{n^2+n-1}{n^2}[/tex] and we want to show this converges to 1. Then we look at the difference |a_n-1|, which simplifies to <br /> |\frac{n-1}{n^2}|=\frac{1}{n}|1-\frac{1}{n}<br /> We only need to make one simplification to find N. We need:<br /> \frac{1}{n}|1-\frac{1}{n}|&amp;lt;h <br /> For any value of n&gt;0, this is always less than<br /> \frac{1}{n}|1|=\frac{1}{n}<br /> Since <br /> \frac{1}{n}|1-\frac{1}{n}|&amp;lt;\frac{1}{n} <br /> Then if<br /> \frac{1}{n}&amp;lt;h<br /> This must mean that<br /> \frac{1}{n}|1-\frac{1}{n}|&amp;lt;h<br /> So we just need to choose n so big that <br /> \frac{1}{n}&amp;lt;h<br /> Choosing N=1/h will make the above inequality true whenever n&gt;N. The above inequality implies that |a_n-1|&lt;h so that&#039;s all you need. You&#039;re problem is very similar from where I left off.

 

[Natasha1]

You have proven that a_n-->1, but you didn't use the definition. Since you've never seen an example of how its done I'll show ou how to do a similar one. suppose a_n=\frac{n^2+n-1}{n^2}[/tex] and we want to show this converges to 1. Then we look at the difference |a_n-1|, which simplifies to <br /> |\frac{n-1}{n^2}|=\frac{1}{n}|1-\frac{1}{n}<br /> We only need to make one simplification to find N. We need:<br /> \frac{1}{n}|1-\frac{1}{n}|&amp;lt;h <br /> For any value of n&gt;0, this is always less than<br /> \frac{1}{n}|1|=\frac{1}{n}<br /> Since <br /> \frac{1}{n}|1-\frac{1}{n}|&amp;lt;\frac{1}{n} <br /> Then if<br /> \frac{1}{n}&amp;lt;h<br /> This must mean that<br /> \frac{1}{n}|1-\frac{1}{n}|&amp;lt;h<br /> So we just need to choose n so big that <br /> \frac{1}{n}&amp;lt;h<br /> Choosing N=1/h will make the above inequality true whenever n&gt;N. The above inequality implies that |a_n-1|&lt;h so that&#039;s all you need. You&#039;re problem is very similar from where I left off.

<br /> <br /> Thanks LeonhardEuler, you are a gentleman!

 

[Natasha1]

I can't do this one :-(

Prove from the definition that a_n=\frac{n^2 + 3n}{n^2 +2}<br />-> 1 as n -> \infty

Right here is my attempt then:

Lets look at the difference |a_n-1|, which simplifies to

|a_n-1|=\frac{n^2 + 3n}{n^2 +2}-1=|\frac{n^2+3n-n^2-2}{n^2+2}|=|\frac{3n-2}{n^2+2}|

This can be re-written as:

|\frac{3n-2}{n^2}|&lt;h

Since n is positive, to find N, we need:

\frac{1}{n}|3-\frac{2}{n}|&lt;h

For any value of n>0, this is always less than

\frac{1}{n}|3|=\frac{2}{n}

Since

\frac{1}{n}|3-\frac{2}{n}|&lt;\frac{1}{n}

Then if

\frac{2}{n}&lt;h

This must mean that

\frac{1}{n}|3-\frac{2}{n}|&lt;h

So we just need to choose n so big that

\frac{2}{n}&lt;h

Choosing N=2/h will make the above inequality true whenever n>N. The above inequality implies that |an-1|<h.

And therefore

a_n=\frac{n^2 + 3n}{n^2 +2}-> 1 as n -> \infty

 

[LeonhardEuler]

Right here is my attempt then:

Lets look at the difference |a_n-1|, which simplifies to

|a_n-1|=\frac{n^2 + 3n}{n^2 +2}-1=|\frac{n^2+3n-n^2-2}{n^2+2}|=|\frac{3n-2}{n^2+2}|

This can be re-written as:

|\frac{3n-2}{n^2}|&lt;h

This isn't actually re-writing the original equation, its another inequality:
We actualy want
|\frac{3n-2}{n^2+2}|&lt;h
Since we have that
|\frac{3n-2}{n^2+2}|&lt;|\frac{3n-2}{n^2}|
then if |\frac{3n-2}{n^2}|&lt;h, |\frac{3n-2}{n^2+2}| must also be less than h.

Since n is positive, to find N, we need:

\frac{1}{n}|3-\frac{2}{n}|&lt;h

For any value of n>0, this is always less than

\frac{1}{n}|3|=\frac{2}{n}

You mean
\frac{1}{n}|3|=\frac{3}{n}

Since

\frac{1}{n}|3-\frac{2}{n}|&lt;\frac{1}{n}

Then if

\frac{2}{n}&lt;h

This must mean that

\frac{1}{n}|3-\frac{2}{n}|&lt;h

So we just need to choose n so big that

\frac{2}{n}&lt;h

Choosing N=2/h will make the above inequality true whenever n>N. The above inequality implies that |an-1|<h.

And therefore

a_n=\frac{n^2 + 3n}{n^2 +2}-> 1 as n -> \infty

The rest is right as long as you fix the problem with the 3.