# Lim convergent sequence

1. Dec 7, 2009

### gtfitzpatrick

1. The problem statement, all variables and given/known data

Proove rigorously that if (a$$_{n}$$ is a real convergent sequence with lim$$_{n\rightarrow \infty}$$ a$$_{n}$$ = a and for each n=$$\in$$ N, a$$_{n}$$ < 6, then a $$\leq$$ 6
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Let $$\epsilon$$ > 0 we need to find n$$_{0}$$ $$\in$$ N such that

$$\left\|$$ a$$_{n}$$ - a$$\left\|$$ < $$\epsilon$$ $$\forall$$ n $$\geq$$ n $$_{0}$$, n$$_{0}$$ $$\in$$ N

but a$$_{n}$$ < 6
so

$$\left\|$$ 6 - a$$\left\|$$ < $$\epsilon$$

then

a < 6 - $$\epsilon$$ and $$\epsilon$$ > 0

so a $$\leq$$ 6

i think i've done this right, just by using the definition of a limit. Could anyone tell me if this is looking ok?

(Also i cant seem to get the sub script working, it always makes them go up instead of down, any ideas anyone?)

Thanks a million

Last edited: Dec 7, 2009
2. Dec 7, 2009

No, you're off track. You know that the sequence converges, so no matter what $$\epsilon$$ is given you can make

$$|a_n - a| < \epsilon$$

if n is large enough.
But this:

doesn't follow.

Try assuming $$a > 6$$ and see if you can reach some contradiction. (Hint: if $$a > 6$$ then $$a - 6 > 0$$.)

3. Dec 8, 2009

### gtfitzpatrick

thanks for getting back to me.

ok let me see if i have this right now.

first since the sequence is convergent $$\epsilon$$ > 0. then set a > 6 so

Since the sequence is convergent a$$_{}n$$ - a < $$\epsilon$$ or a < a$$_{}n$$ - $$\epsilon$$ but if a > 6 then 6 - $$\epsilon$$ > a > 6 which cant be true so a muxt be < or = 6

4. Dec 8, 2009

Not quite . My point was that since $$a - 6 >0$$ it would be a possibility for choice as $$\epsilon$$

5. Dec 8, 2009

### gtfitzpatrick

thanks for all the help,

do you mean a - 6 > 0 and an -a < $$\epsilon$$ then an - a < a - 6 ?

6. Dec 8, 2009

### gtfitzpatrick

I must be still lookin at this wrong, cant seem to figure it out

7. Dec 8, 2009

### JG89

Think of it intuitively first.

If $$a_n < 6$$ for all n, and you want to prove that $$a \le 6$$, then assume the opposite. Assume that $$a > 6$$. But since $$a_n \rightarrow a$$ then that means we can make $$a_n$$ as close to a as we want, provided that n is taken large enough, right?

Do you see how if a > 6 then there is no way we can possibly make $$a_n$$ as close to a as we want, no matter how large n is? Because for all n, $$a_n < 6$$ ?

Make this into a rigorous argument now.

8. Dec 8, 2009

### gtfitzpatrick

I can see that if an < 6 and a > 6 but an $$\rightarrow$$ a so they cant converge as they are both on opposite sides of 6 so to speak, but im just not sure how to go about prooving this rigourously

9. Dec 8, 2009

### JG89

Assume a > 6. Since $$a_n < 6 < a$$ for all n, we have $$a - a_n > 0$$. So $$|a_n - a| = a - a_n$$.

Now, we know that for every positive $$\epsilon$$ , $$a - a_n < \epsilon$$ provided that n is large enough. If a > 6 then a - 6 > 0, and so we can take $$\epsilon = a - 6$$. Try that out and see what happens.

10. Dec 8, 2009

### gtfitzpatrick

the first part of this is grand. the second part, im not following are you getting it from the def of a limit $$\left|$$an-a $$\left|$$ < $$\epsilon$$

11. Dec 8, 2009

### JG89

In the part you quoted, I'm just making it clear that |a_n - a| = a - a_n.

I'd rather get rid of the absolute value bars.

The rest of my post uses the definition of a limit.

12. Dec 8, 2009

### gtfitzpatrick

sorryi'm not getting this, is this just because a>6>a_n

13. Dec 8, 2009

### JG89

Yes. You know that |a_n - a| = |a - a_n|, right? but a - a_n > 0 anyway since a > a_n, so we can just drop the absolute value bars and write a - a_n

14. Dec 8, 2009

### gtfitzpatrick

Damn this just isnt working for me- $$\epsilon = a- 6$$ but $$\epsilon>0$$ so a-6> 0 or a>6 which is what i'm trying to disprove!

15. Dec 8, 2009

### gtfitzpatrick

a-an < $$\epsilon$$ and a - 6 = $$\epsilon$$

then a-an < a - 6

which works out to an < 6 which is true???

16. Dec 8, 2009

$$a - a_n < a - 6$$

$$a_n < 6$$?