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

<br /> |a_n - a| &lt; \epsilon<br />

if n is large enough.
But this:

but a_{n} < 6
so

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

doesn't follow.

Try assuming a &gt; 6 and see if you can reach some contradiction. (Hint: if a &gt; 6 then a - 6 &gt; 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 &gt;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 &lt; 6 for all n, and you want to prove that a \le 6, then assume the opposite. Assume that a &gt; 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 &lt; 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 &lt; 6 &lt; a for all n, we have a - a_n &gt; 0. So |a_n - a| = a - a_n.

Now, we know that for every positive \epsilon , a - a_n &lt; \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 &lt; 6 &lt; a for all n, we have a - a_n &gt; 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&gt;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?

<br /> a - a_n &lt; a - 6<br />

leads to

<br /> a_n &lt; 6<br />?

Check your signs again.

 

[gtfitzpatrick]

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