Convergence of Sequence Proof: Is This Correct?

[Mr Davis 97]

Homework Statement

Prove rigorously that $\displaystyle \lim \frac{n}{n^2 + 1} = 0$.

Homework Equations

A sequence $(s_n)$ converges to $s$ if $\forall \epsilon > 0 \exists N \in \mathbb{N} \forall n \in \mathbb{N} (n> N \implies |s_n - s| < \epsilon)$

The Attempt at a Solution

Let $\epsilon > 0$. Let $\displaystyle N = \frac{1}{\epsilon}$. Let $n \in \mathbb{N}$.
Then, if $n > N$, we have that $\displaystyle n > \frac{1}{\epsilon}$ and so $\displaystyle \frac{1}{n} < \epsilon$. Therefore, $\displaystyle |\frac{n}{n^2+1} - 0| = \frac{n}{n^2+1} < \frac{n}{n^2} = \frac{1}{n} < \epsilon$. This proves that $\displaystyle \lim \frac{n}{n^2 + 1} = 0$.

Is this a correct convergence proof?

 

[LCKurtz]

Looks OK to me.

 

[Mr Davis 97]

Looks OK to me.

I actually have a concern. I let $\displaystyle N = \frac{1}{\epsilon}$, but by the definition of sequence convergence, shouldn't $N$ be an integer? Is this a problem? How could I fix it?

 

[Mark44]

I actually have a concern. I let $\displaystyle N = \frac{1}{\epsilon}$, but by the definition of sequence convergence, shouldn't $N$ be an integer? Is this a problem? How could I fix it?

Take N = $\lceil \frac 1 \epsilon \rceil$, IOW, the smallest integer greater than or equal to $\frac 1 \epsilon$.

 

[LCKurtz]

I actually have a concern. I let $\displaystyle N = \frac{1}{\epsilon}$, but by the definition of sequence convergence, shouldn't $N$ be an integer? Is this a problem? How could I fix it?

Nothing requires N to be an integer. It is the n > N that is an integer.

 

[Mr Davis 97]

Nothing requires N to be an integer. It is the n > N that is an integer.

Well my definition claims that $N$ is a natural number. Would Mark44's approach suffice?

 

[LCKurtz]

I don't know about your textbook, but it doesn't matter whether N is an integer. Here's the definition of convergence from another source:
http://mathworld.wolfram.com/ConvergentSequence.html

 

[Mark44]

@Mr Davis 97, you're overthinking this. If $\frac 1 \epsilon$ isn't an integer, just take N to be the next-larger integer, which is basically what I said before. After all, N is just a lower bound on the indexes of the sequence elements that are "small enough." Elements further in the sequence will be even smaller; i.e., closer to zero.