Proving the Convergence of (2n^2+n)/(n^2) to 2

[jeff1evesque]

I am trying to prove a larger problem, that the sequence (2n^2+n)/(n^2) --> 2

however, i need something small to prove it which is proving the fact that given n > 2,

2^n - n > n

THanks,


JL

 

[sutupidmath]

Try induction!

 

[jeff1evesque]

Try induction!

Exactly what i was thinking, but i need it in a different form to fit my proof.

Proof:
Let e > 0 be given.
Let N = max{2, 1/n}
Note that, |1/(2^n - n) - 0| = |1/(2^n - n)| (trying to show later in the proof this is less than 1/n)

Assume n > N
then n > 2 (Goal: show that 1/(2^n - n) < (1/n) )
.
.
.
then since |xn - L| < 1/n
.
.
.

 

[jeff1evesque]

I need to show 2^n - n > n

But I need to start the proof with the assumption that n > 2.

 

[sutupidmath]

I need to show 2^n - n > n

But I need to start the proof with the assumption that n > 2.

you can still use induction to prove it? What do you think prevents you from using induction here?

 

[sutupidmath]

Contents of deleted post[/color]

he/she probbably needs to prove it using epsilond delta.(formal proof)

 

[jeff1evesque]

he/she probbably needs to prove it using epsilond delta.(formal proof)

Yeah that's it. Except i was just given sequence (2n^2+n)/(n^2) --> 2. And in my idea stage, I have the sequence equal to 1/(2^n - n) < 1/n.. and this is only true if n > 2.

 

[sutupidmath]

Yeah that's it. Except i was just given sequence (2n^2+n)/(n^2) --> 2. And in my idea stage, I have the sequence equal to 1/(2^n - n) < 1/n.. and this is only true if n > 2.

you seem to be almost done. where is the problem here! it doesn't have to start with n=1 for you to apply induction, as a matter of fact induction holds even if you start at any k positive integer.

 

[statdad]

Yeah that's it. Except i was just given sequence (2n^2+n)/(n^2) --> 2. And in my idea stage, I have the sequence equal to 1/(2^n - n) < 1/n.. and this is only true if n > 2.

If your initial sequence involves only $$n^2$$, how do you arrive at a a statement that has $$2^n$$ involved?

As I read it, your initial sequence is

$$\frac{2n^2 + n}{n^2}$$

If this is correct, just simplify the fraction that gives the nth term and you are 99% of the way to proving the sequence converges to 2. If it isn't correct, you have a typo in your original post on the form of the sequence.