# Epsilon-N Proof of Sequence

1. Oct 28, 2009

### roam

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

Given $$\epsilon >0$$, we can find $$n \geq N$$ such that:

$$| \frac{1}{1+n+n^4} -0 | = \frac{1}{1+n+n^4} < \frac{1}{1+n}< \epsilon$$

Now what value of N should we take to complete the proof? And why?

This is what I guess:
We have $$\frac{1}{1+n}< \epsilon$$ so,

$$n+1> \frac{1}{\epsilon}$$

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

$$N = \frac{1}{\epsilon} -1$$

Is this right? I appreciate it if anyone could provide me with some explaination.

2. Oct 28, 2009

### lanedance

i think that will work, in that for the N you give you know every term with n>N will be less than epsilon.... however, i think you can put much tighter constraints on N using the n^4 part of the sequence.

1/n^4 goes to zero a lot quicker than 1/n. And for n>>1, n^4 + n + 1 looks more like n^4 than n