Thread: Sequence

I can't do this one :-(

Prove from the definition that -> 1 as 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 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.

Haven't got a clue, please help.

Originally Posted by Natasha1

I can't do this one :-(

Prove from the definition that -> 1 as 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 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.
Haven't got a clue, please help.

Hello,

I haven't much time now, so I show you the steps without much explanation:
Calculate the difference:
Use synthetic division.







Greetings

EB

Originally Posted by Natasha1

I can't do this one :-(

Prove from the definition that -> 1 as 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 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.

Haven't got a clue, please help.

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"?

How about this way.
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.

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