1. ## Sequence

I can't do this one :-(

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

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 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.

2. Originally Posted by Natasha1
I can't do this one :-(

Prove from the definition that $\displaystyle a_n=\frac{n^2 + 3n}{n^2 +2}$ -> 1 as n -> $\displaystyle \infty$
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 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.
Hello,

I haven't much time now, so I show you the steps without much explanation:
Calculate the difference:
$\displaystyle \frac{n^2 + 3n}{n^2 +2} -1 \leq h$ Use synthetic division.

$\displaystyle 1+\frac{3n -2}{n^2 +2} -1 \leq h$

$\displaystyle \frac{n(3 -{2\over n})}{n(n +{2\over n})}\leq h$

$\displaystyle \frac{3 }{n}\leq h$

Greetings

EB

3. Originally Posted by Natasha1
I can't do this one :-(

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

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 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.

Definition? What should be the definition here? I am lost.

Suppose we just show that a_n tends to 1 as n approaches infinity, without using that "definition"?

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

Shown.

4. Originally Posted by ticbol
Definition? What should be the definition here? ...
Hello,

I guessed that Natasha1 refered to Cauchy's convergence criteria.

May be I was wrong, but finally I got the simplier expression which was asked for.

Greetings

EB