Another Epsilon-N Limit Proof Question

Question

How to prove the limit of the following sequence using epsilon-N argument.

$$a_n=\frac{3n^2+2n+1}{2n^2+n}$$

I took the limit to be $\frac{3}{2}$ and proceeded with the argument,

$$\left|\frac{3n^2+2n+1}{2n^2+n}-\frac{3}{2}\right|<\epsilon$$ $$\frac{n+2}{4n^2+2n}<\epsilon$$ How do I complete the argument from this point?

Answer 1

$$\frac{n+2}{4n^2 + 2n} = \frac{n+2}{2n(2n+1)} \leq \frac{n+2}{2n(n+2)} = \frac{1}{2n}$$ Do you want to take it from here? Maybe you'd like to see a full $\epsilon - N$ proof?

Answer 2

When you get an expression of the form $f(n)<\varepsilon$, what you'd like to do is rewrite the inequality in terms of $\varepsilon$, that is, you'd like to write $g(\varepsilon )<n$, for some $g$. Then you take $n$ a natural number that satisfies this inequality and you're done.

Often times $f(n)$ is not simple enough (sometimes even impossible) to do this, so you need to find another function $h$ for the same purpose (rewriting $h(n)<\varepsilon$ with respect to $\varepsilon$) with the additional properties that $f(n)<h(n)$ and $\lim \limits_{n\to +\infty}\left(h(n)\right)=0$.

It's easy to see that $\dfrac{n+2}{4n^2+2n}<\dfrac{n+2}{4n^2}$.

Let $h(n)=\dfrac {n+2}{4n^2}$ and consider $h(n)<\varepsilon$.

Now consider $h(n)=\varepsilon$. This is equivalent to $4n^2\varepsilon -n-2=0$ which in turn is equivalent to $n=\dfrac{1\pm \sqrt{1+32\varepsilon}}{8\varepsilon}$, so the inequality $h(n)<\varepsilon$ will be satisfied just as long as $n$ is a natural number greater than $\dfrac{1+ \sqrt{1+32\varepsilon}}{8\varepsilon}$.

Getting back to your proof, you wish to prove that $$\forall \varepsilon >0\exists N\in \mathbb N \forall n\in \mathbb N\left(n\ge N\implies \left|\frac{3n^2+2n+1}{2n^2+n}-\frac{3}{2}\right|<\varepsilon\right).$$

Take $\varepsilon>0$ and let $N$ be any natural number greater than $\dfrac{1+ \sqrt{1+32\varepsilon}}{8\varepsilon}$. Take any natural number $n$ greater than $N$.

Your goal is to prove that $$\left|\frac{3n^2+2n+1}{2n^2+n}-\frac{3}{2}\right|<\varepsilon,$$

which is equivalent to proving $$\frac{n+2}{4n^2+2n}<\epsilon.$$

You know that $N$ satisfies $f\left(N\right)<h\left(N\right)<\varepsilon$, i.e., $N$ satisfies the inequality above. It's easy to prove that because $n\ge N$, then $n$ will satisfy $f(n)<f\left(N\right)$ and consequently it will satisfy the inequality $\left|\frac{3n^2+2n+1}{2n^2+n}-\frac{3}{2}\right|<\varepsilon$, as is needed.

Of course Dror's $h$ in his answer is much, much simpler, but my remarks are still compatible with his choice for $h$