# [SOLVED] Epsilon-N Proof

• Oct 28th 2009, 04:45 PM
Roam
[SOLVED] Epsilon-N Proof
Prove that the sequence $\displaystyle (\frac{1}{1+n+n^4})$ converges to 0.

Here's the solution:
http://img517.imageshack.us/img517/7235/81768116.gif

Here's my problem: I don't understand why they wrote $\displaystyle N=\frac{1}{\epsilon}$ !

We know that $\displaystyle n+1 > \frac{1}{\epsilon}$

Therefore $\displaystyle n> \frac{1}{\epsilon} -1$

And so we should take: $\displaystyle N= \frac{1}{\epsilon} -1$

Isn't that right???

Also, I don't understand why they first omitted $\displaystyle n^4$ from the denominator when they could omit $\displaystyle n+1$ instead. Because I think $\displaystyle \frac{1}{n^4}$ goes to zero a lot quicker. I appreciate some explanation.
• Oct 28th 2009, 05:22 PM
proscientia
Quote:

Originally Posted by Roam
We know that $\displaystyle n+1 > \frac{1}{\epsilon}$

Therefore $\displaystyle n> \frac{1}{\epsilon} -1$

And so we should take: $\displaystyle N= \frac{1}{\epsilon} -1$

No. Firstly, $\displaystyle N$ must be an integer. $\displaystyle \frac1\epsilon-1$ may not be an integer. Secondly, $\displaystyle n+1>\frac1\epsilon$ is what we want, not what we know. By choosing $\displaystyle N=\left\lceil\frac1\epsilon\right\rceil,$ we can be assured that whenever $\displaystyle n>N,$ we have $\displaystyle n+1>\frac1\epsilon.$
• Oct 29th 2009, 12:54 AM
Roam
Quote:

Originally Posted by proscientia
$\displaystyle \frac1\epsilon-1$ may not be an integer.

Why is that?

And, what about my second question? Why couldn't we start out by omitting $\displaystyle 1+n$ instead of $\displaystyle n^4$?

$\displaystyle \frac{1}{1+n+n^4} < \frac{1}{n^4} < \epsilon$

Thus $\displaystyle N= \frac{1}{\epsilon ^4}$
• Oct 29th 2009, 02:44 AM
Plato
Quote:

Originally Posted by Roam
Thus $\displaystyle N= \frac{1}{\epsilon ^4}$

That does not work.
There is no reason that $\displaystyle \frac{1}{\epsilon ^4}$ is an integer.
• Oct 29th 2009, 10:30 AM
Roam
Quote:

Originally Posted by Plato
That does not work.
There is no reason that $\displaystyle \frac{1}{\epsilon ^4}$ is an integer.

Then how do you know that $\displaystyle \frac{1}{\epsilon}$ is an integer?
• Oct 29th 2009, 02:35 PM
proscientia
Where is it stated that $\displaystyle \frac1\epsilon$ is an integer? $\displaystyle N$ is $\displaystyle \left\lceil\frac1\epsilon\right\rceil,$ not $\displaystyle \frac1\epsilon.$
• Oct 30th 2009, 06:41 AM
HallsofIvy
Quote:

Originally Posted by Roam
Then how do you know that $\displaystyle \frac{1}{\epsilon}$ is an integer?

You don't! If, for example, $\displaystyle \epsilon= .00000003= \frac{3}{100000000}$ then $\displaystyle \frac{1}{\epsilon}= \frac{100000000}{3}$ which is not an integer.

If you look in any text book you should see that they say something like $\displaystyle N> \frac{1}{\epsilon} -1$, not $\displaystyle N= \frac{1}{\epsilon} -1$