1. ## [SOLVED] Epsilon-N Proof

Prove that the sequence $(\frac{1}{1+n+n^4})$ converges to 0.

Here's the solution:

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

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

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

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

Isn't that right???

Also, I don't understand why they first omitted $n^4$ from the denominator when they could omit $n+1$ instead. Because I think $\frac{1}{n^4}$ goes to zero a lot quicker. I appreciate some explanation.

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

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

And so we should take: $N= \frac{1}{\epsilon} -1$
No. Firstly, $N$ must be an integer. $\frac1\epsilon-1$ may not be an integer. Secondly, $n+1>\frac1\epsilon$ is what we want, not what we know. By choosing $N=\left\lceil\frac1\epsilon\right\rceil,$ we can be assured that whenever $n>N,$ we have $n+1>\frac1\epsilon.$

3. Originally Posted by proscientia
$\frac1\epsilon-1$ may not be an integer.
Why is that?

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

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

Thus $N= \frac{1}{\epsilon ^4}$

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

5. Originally Posted by Plato
That does not work.
There is no reason that $\frac{1}{\epsilon ^4}$ is an integer.
Then how do you know that $\frac{1}{\epsilon}$ is an integer?

6. Where is it stated that $\frac1\epsilon$ is an integer? $N$ is $\left\lceil\frac1\epsilon\right\rceil,$ not $\frac1\epsilon.$

7. Originally Posted by Roam
Then how do you know that $\frac{1}{\epsilon}$ is an integer?
You don't! If, for example, $\epsilon= .00000003= \frac{3}{100000000}$ then $\frac{1}{\epsilon}= \frac{100000000}{3}$ which is not an integer.

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