# Basic analysis

1. Feb 11, 2010

### 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

2. Feb 11, 2010

### sutupidmath

Try induction!

3. Feb 11, 2010

### jeff1evesque

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
.
.
.

4. Feb 11, 2010

### jeff1evesque

I need to show 2^n - n > n

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

5. Feb 11, 2010

### sutupidmath

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

6. Feb 11, 2010

### sutupidmath

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

Last edited by a moderator: Feb 11, 2010
7. Feb 11, 2010

### jeff1evesque

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.

8. Feb 11, 2010

### sutupidmath

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.

9. Feb 12, 2010

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