Proving Convergence of a Sequence with a Given Upper Bound

[gtfitzpatrick]

Homework Statement

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

Homework Statement

Homework Equations

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 can't seem to get the sub script working, it always makes them go up instead of down, any ideas anyone?)

Thanks a million

 

[statdad]

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:

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

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

doesn't follow.

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

 

[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 can't be true so a muxt be < or = 6

 

[statdad]

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

 

[gtfitzpatrick]

thanks for all the help,

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

 

[gtfitzpatrick]

I must be still lookin at this wrong, can't seem to figure it out

 

[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.

 

[gtfitzpatrick]

thanks a mill for replying

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

 

[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.

 

[gtfitzpatrick]

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

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

 

[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.

 

[gtfitzpatrick]

|a_n - a| = a - a_n.

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

 

[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

 

[gtfitzpatrick]

Damn this just isn't working for me- $$\epsilon = a- 6$$ but $$\epsilon>0$$ so a-6> 0 or a>6 which is what I'm trying to disprove!

 

[gtfitzpatrick]

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

then a-a_n < a - 6

which works out to a_n < 6 which is true?

 

[statdad]

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

then a-a_n < a - 6

which works out to a_n < 6 which is true?

$$a - a_n < a - 6$$

leads to

$$a_n < 6$$?

Check your signs again.

 

[gtfitzpatrick]

damn it, sorry bout that. thanks a million for the help