# Help with Intro to Real Analysis proof with sequence and limit.

• Nov 1st 2008, 08:43 AM
ilikedmath
Help with Intro to Real Analysis proof with sequence and limit.
Problem states:
If a > 0, prove http://www.cramster.com/Answer-Board...0937504331.gif (the set of Natural numbers) such that http://www.cramster.com/Answer-Board...6875002318.gif.

My work so far:
I don't have a proof written up for this problem yet, but I have jotted down some ideas that I think might help me:

I know that the sequence {$\displaystyle a_n$} converges to a and a > 0.
So by definition of convergence, I know that there exists an $\displaystyle n_0$ in the Naturals such that
|$\displaystyle a_n$ - a| < ε for all n ≥ $\displaystyle n_0$.

There is also the theorem that every convergent sequence is bounded, so by definition of bounded: there exists an M > 0 such that
|$\displaystyle a_n$| ≤ M for all n in the Naturals.

So for this proof do I need cases? Would I do proof by contradiction?
I worked with a study group on this, and my classmate drew a picture and talked us through the thinking. It kind of made sense to me that if a sequence converges to zero, then eventually the sequence has to be greater than zero at some point.

Any help, suggestions, and/or tips are greatly appreciated.
Thank you for your time! =)
• Nov 1st 2008, 10:18 AM
Plato
From the fact that $\displaystyle \left( {a_n } \right) \to a>0$ you use the definition of sequence convergence.
But in place of just $\displaystyle \varepsilon >0$ you are free to use $\displaystyle \frac{a}{2} >0$.
$\displaystyle \begin{gathered} \left| {a_n - a} \right| < \frac{a}{2} \hfill \\ - \frac{a}{2} < a_n - a < \frac{a}{2} \hfill \\ 0 < \frac{a}{2} < a_n < \frac{{3a}}{2} \hfill \\ \end{gathered}$
• Nov 1st 2008, 10:26 AM
ilikedmath
Okay, so can I just choose any value for epsilon? Like could I have chosen a/3 or a/7 > 0?

Here's my proof so far:
http://i35.tinypic.com/166xhsh.jpg
QED.

is that okay?
• Dec 21st 2008, 09:06 PM
Mathstud28
You could do this proof by contradiction. Assume that there does not exist a $\displaystyle N$ such that $\displaystyle N\leqslant n\implies a_n>0$

By definition of $\displaystyle \lim_{n\to\infty}a_n=a$ we must have that for every $\displaystyle \varepsilon>0$ there exists a corresponding $\displaystyle N$ such that $\displaystyle N\leqslant n\implies|a_n-a|<\varepsilon$. So if we take $\displaystyle \varepsilon=\frac{a}{2}$ we must be able to find a $\displaystyle N$ such that $\displaystyle |a_n-a|<\frac{a}{2}$, but now by assumption $\displaystyle a\leqslant|a_n-a|$ which gives a contradiction.
• Dec 22nd 2008, 07:36 AM
CaptainBlack
Quote:

Originally Posted by Mathstud28
You could do this proof by contradiction. Assume that there does not exist a $\displaystyle N$ such that $\displaystyle N\leqslant n\implies a_n>0$

By definition of $\displaystyle \lim_{n\to\infty}a_n=a$ we must have that for every $\displaystyle \varepsilon>0$ there exists a corresponding $\displaystyle N$ such that $\displaystyle N\leqslant n\implies|a_n-a|<\varepsilon$. So if we take $\displaystyle \varepsilon=\frac{a}{2}$ we must be able to find a $\displaystyle N$ such that $\displaystyle |a_n-a|<\frac{a}{2}$, but now by assumption $\displaystyle a\leqslant|a_n-a|$ which gives a contradiction.

Confused, it can be done this way (though I can't see why one would want to) and this is probably OK when all the missing detail is put in but in this form it is near incomprehensible (and the last $\displaystyle n$ is not the same as the $\displaystyle n$ introduced earlier).

CB
• Dec 22nd 2008, 03:23 PM
Mathstud28
Quote:

Originally Posted by CaptainBlack
Confused, it can be done this way (though I can't see why one would want to) and this is probably OK when all the missing detail is put in but in this form it is near incomprehensible (and the last $\displaystyle n$ is not the same as the $\displaystyle n$ introduced earlier).

CB

Yes you are right, that is unclear. Let me fix it up.

Suppositions:
1. $\displaystyle a>0$
2. $\displaystyle \lim a_n=a$
3. There exists a $\displaystyle N$ such that $\displaystyle N\leqslant n\implies a_n\leqslant 0$

A fact we will use, if $\displaystyle x>0$ and $\displaystyle y<0$ then $\displaystyle x\leqslant|x-y|$

Now by the definition of a sequence $\displaystyle \left\{a_n\right\}$ converging to $\displaystyle a$ we must have that for every $\displaystyle \varepsilon>0$ there exists a corresponding $\displaystyle N_1$ such that $\displaystyle N_1\leqslant n\implies |a-a_n|<\frac{a}{2}$. So now let $\displaystyle \varepsilon=\frac{a}{2}$, so we must be able to find a $\displaystyle N_1$ such that $\displaystyle N_1<n\implies |a_n-a|<\varepsilon$. So it must be true that if we take $\displaystyle N_2=\max\left\{N,N_1\right\}$ that $\displaystyle N_2\leqslant n\implies |a_n-a|<\frac{a}{2}$. But here we arrive at a contradiction. The reason being that because of supposition three we must have that $\displaystyle N_2\leqslant n\implies |a-a_n|\geqslant a\implies |a_n-a|\not<\frac{a}{2}$