Does This Epsilon-N Proof Correctly Show a Sequence's Convergence to Zero?

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SUMMARY

The discussion centers on proving the convergence of the sequence \( \frac{1}{1+n+n^4} \) to 0. The proposed solution involves finding a suitable \( N \) such that for all \( n \geq N \), \( \left| \frac{1}{1+n+n^4} - 0 \right| < \epsilon \). The conclusion suggests that setting \( N = \frac{1}{\epsilon} - 1 \) is valid, but a more precise approach is recommended by considering the dominant term \( n^4 \) in the denominator, which leads to a faster convergence rate. This indicates that tighter constraints on \( N \) can be established using the \( n^4 \) term.

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Homework Statement



Prove that the sequence [tex](\frac{1}{1+n+n^4})[/tex] converges to 0.


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The Attempt at a Solution



Given [tex]\epsilon >0[/tex], we can find [tex]n \geq N[/tex] such that:

[tex]| \frac{1}{1+n+n^4} -0 | = \frac{1}{1+n+n^4} < \frac{1}{1+n}< \epsilon[/tex]

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

This is what I guess:
We have [tex]\frac{1}{1+n}< \epsilon[/tex] so,

[tex]n+1> \frac{1}{\epsilon}[/tex]

[tex]n>\frac{1}{\epsilon} -1[/tex]

[tex]N = \frac{1}{\epsilon} -1[/tex]

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

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