Proving Limit with Epsilon and Delta

[msell2]

Homework Statement

Prove the following sequence {a_n} converges to L=1/2
a_n = n²/(2n²+n-1)

The Attempt at a Solution

Given ε>0 we can determine an N∈N so that |a_n - L|<ε for n≥N. We have:
|a_n-L|=|(n²/(2n²+n-1)-(1/2)|
= |(-n+1)/(2(2n-1)(n+1))|

I'm not sure what to do once I get to this point. Any help would be great!

 

[LCKurtz]

Homework Statement

Prove the following sequence {a_n} converges to L=1/2
a_n = n²/(2n²+n-1)

The Attempt at a Solution

Given ε>0 we can determine an N∈N so that |a_n - L|<ε for n≥N. We have:
|a_n-L|=|(n²/(2n²+n-1)-(1/2)|
= |(-n+1)/(2(2n-1)(n+1))|

I'm not sure what to do once I get to this point. Any help would be great!

If you change the numerator to (+n+1) you can get that fraction to be less or equal to$$
\frac 1 {2(2n-1)}$$What happens if you drop the $-1$ too?

[Edit] Ignore the drop the -1 part. Maybe change the -1 to -n?

 

[vela]

In a similar vein, you might want to show that $0<\frac{n-1}{2(2n-1)}<1$ for n>1.

 

[PeroK]

The point behind these ideas is that it can be difficult to find N for a given ε unless the expression is quite simple, so you estimate the expression you have with something simpler.

E.g. suppose you wanted to show that 1/(n^2 + 6) -> 0. One way to do this is to note that:

0 < 1/(n^2+6) < 1/n

So, you can take N = 1/ε. Which is much simpler than trying to find a more precise N for the original sequence.

For your example, you want to find a simple expression that is greater than the one you have but still -> 0. That's the strategy, anyway. (See Vela's idea for how to do this.)

 

[msell2]

Thanks for the help! Does this look right?

Given ε>0 we can determine an N\inN so that |a_n-L|<ε for n≥N. We have:
|a_n-L|=|\frac{n^{2}}{2n^{2}+n-1}-\frac{1}{2}|=|\frac{2n^{2}}{2(2n^{2}+n-1}-\frac{2n^{2}+n-1}{2(2n^{2}+n-1}|=|\frac{-n+1}{2(2n^{2}+n-1}|=|\frac{-n+1}{2(2n-1)(n+1)}|<ε.
Now take \frac{n-1}{2(2n-1)}. Since n-1>1 when n>1 and since 4n-2 > n-1 > 1 \Rightarrow 0<\frac{n-1}{2(2n-1)}<1 when n>1. As long as we have n>1, we have that |\frac{-n+1}{2(2n-1)(n+1)}|<\frac{n-1}{2(2n-1)}. We want to bound this by ε>0. Take ε>0. If we choose N to be the smallest integer greater than or equal to max{1,1/ε} then we have that, for n≥N:
|\frac{n^{2}}{2n^{2}+n-1}-\frac{1}{2}|<\frac{1}{n}<\frac{1}{N}≤ε
where the last line follows directly if N≥1/ε and is also satisfied for N=1>1/ε since this implies 1/N<ε.

 

[vela]

Thanks for the help! Does this look right?

Given ε>0 we can determine an N\inN so that |a_n-L|<ε for n≥N. We have:
|a_n-L|=|\frac{n^{2}}{2n^{2}+n-1}-\frac{1}{2}|=|\frac{2n^{2}}{2(2n^{2}+n-1}-\frac{2n^{2}+n-1}{2(2n^{2}+n-1}|=|\frac{-n+1}{2(2n^{2}+n-1}|=|\frac{-n+1}{2(2n-1)(n+1)}|<ε.
Now take \frac{n-1}{2(2n-1)}. Since n-1>1 when n>1 and since 4n-2 > n-1 > 1 \Rightarrow 0<\frac{n-1}{2(2n-1)}<1 when n>1. As long as we have n>1, we have that |\frac{-n+1}{2(2n-1)(n+1)}|<\frac{n-1}{2(2n-1)}.

Is that last inequality what you intended?

We want to bound this by ε>0. Take ε>0. If we choose N to be the smallest integer greater than or equal to max{1,1/ε} then we have that, for n≥N:
|\frac{n^{2}}{2n^{2}+n-1}-\frac{1}{2}|<\frac{1}{n}<\frac{1}{N}≤ε
where the last line follows directly if N≥1/ε and is also satisfied for N=1>1/ε since this implies 1/N<ε.

 

[msell2]

What is wrong with that inequality?

 

[vela]

How are you going from

\lvert \frac{-n+1}{2(2n-1)(n+1)} \rvert &lt; \frac{n-1}{2(2n-1)}

to

We want to bound this by ε>0. Take ε>0. If we choose N to be the smallest integer greater than or equal to max{1,1/ε} then we have that, for n≥N:
\lvert \frac{n^{2}}{2n^{2}+n-1}-\frac{1}{2}\rvert &lt; \frac{1}{n} &lt; \frac{1}{N}≤\varepsilon
where the last line follows directly if N≥1/ε and is also satisfied for N=1>1/ε since this implies 1/N<ε.

 

[dirk_mec1]

n-1< 4n -2 => 1<3n which is true for n>1 get it?