Prove convergence using $\varepsilon$-$N$ definition

This is an example presented by professor in class. I understand the idea behind this kind of definition, but I'm having trouble following my professor's thought process.

We must prove that the $\displaystyle \lim_{n\to\infty} \frac{2n-1}{n^2} = 0$ using the $\varepsilon$-$N$ definition of a limit of a sequence.

So, the foregoing limit is true iff $\forall \varepsilon>0,\ \exists N(\varepsilon)>0 : \forall n \ ( n \geq N \Rightarrow |f(n) - 0| < \varepsilon)$.

Now, the way I understand it is that if we prove the conditional statement by choosing an appropriate $N$, we've proven the limit.

So first we fix $\varepsilon>0$ since it's given, and play around with $$\displaystyle \left| \frac{2n-1}{n^2} \right| < \varepsilon$$ $$|2n-1| < \varepsilon \ n^2 .$$

We note that $|2n-1|<|2n|$ which sits right with me.

However, my professor then reasons that $|2n-1|<|2n|<\varepsilon \ n^2$ and so $2n < \varepsilon \ n^2$.

How do we know that $2n < \varepsilon \ n^2$? If epsilon is really small, wouldn't there be a point when this inequality is no longer true? This is what is bothering me.

From that we conclude that $\displaystyle n>\frac{\varepsilon}{2}$ and choose $\displaystyle N=\frac{\varepsilon}{2}$ so that

$$n \geq \frac{\varepsilon}{2} \Rightarrow \left| \frac{2n-1}{n^2} \right| < \varepsilon$$

is always true. And that is true iff the limit is true, so the limit is true.

I'd like to know if my understanding of this definition is correct, and if anyone can explain to me the reasoning behind the inequalities above.

Thank you very much!

• The point is that because $2n<\varepsilon n^2\iff 2<\varepsilon n$, no matter how small $\varepsilon$ is, you can find an $n$ such that the inequality will be true, (and more importantly it will be true for $n, n+1, n+2, \ldots$. Does this help? Oct 19, 2013 at 18:15
• Oh yes! Just like our original inequality is contrived, we still can find an $n$ large enough for it to be true, and then let that large $n$ be $N$. So we simply introduce $|2n|$, assume it to be less than $\varepsilon n^2$, and instead use that to find a large $N$ since $|2n-1|<|2n|$. Correct? Oct 19, 2013 at 18:33
• Spot on.${{{}}}$ Oct 19, 2013 at 18:39

I'll give you another way to think about it. Instead of putting $\epsilon$ there from beginning, my approach is always to simply find an upper bound for $|a_n - L|$, and then look at it to figure it out which conditions put over $n$ assure that this bound is less than $\epsilon$.

Your sequence is $(a_n)$ where $a_n = (2n-1)/n^2$. Now, we want to bound $|a_n - 0|$ and look at that bound to see what $n$ must be bigger than in order to make that less than $\epsilon$.

Since $n \geq 1$ we have that $2n - 1 > 0$ and so $|a_n| = a_n$, in other words, we can look just to $(2n-1)/n^2$. Now, this is the same as

$$\dfrac{2n-1}{n^2}=2\dfrac{1}{n}-\dfrac{1}{n^2},$$

Now, look that $-1/n^2 < 0$ always, so that

$$\dfrac{2n-1}{n^2} <\dfrac{2}{n}$$

Now, what must $n$ be greather than in order to make that less than $\epsilon$. Of course, if $n > 1/\epsilon$, then $1/n < \epsilon$, however there's a $2$ there, so that to take care of it we make $n > 2/\epsilon$, now when we divide by $n$ we get $1/n < \epsilon/2$, and hence the condition follows. So, given $\epsilon > 0$, if $n > 2/\epsilon$ we have that

$$\left|\dfrac{2n-1}{n^2}-0\right|=\dfrac{2n-1}{n^2}=\dfrac{2}{n}-\dfrac{1}{n^2}<\dfrac{2}{n} <\epsilon$$

and hence $a_n \to 0$ when $n\to \infty$.

• Right so we simply set the simpler expression, in this case the upper bound $\frac{2}{n}$, to be less than epsilon, because that will make $\frac{2n-1}{n^2} < \epsilon$. Then once we solve for $n$ in terms of $\epsilon$, we know that is what $n$ should be given some $\epsilon$ to make the inequality true. In this case your $N$ would be $\frac{2}{\epsilon}$, which is interesting to see, since it's different than the $N$ in the original post. My understanding of this has definitely improved. Thank you. Oct 19, 2013 at 19:11
• Guys I think this proof is quite simple if you use the comparison lemma: by noting $0<\frac{2n-1}{n^2} = \frac{2n}{n^2}-\frac{1}{n^2}= \frac{2}{n}-\frac{1}{n^2} < \frac{2}{n}$. Since $|\frac{2n-1}{n^2}-0| \le 2|\frac{1}{n}-0| \,\, \forall n \ge 1$, then by setting $N=1\in\mathbb{N}$, then since $\frac{1}{n} \rightarrow 0$ then $\frac{2n-1}{n^2} \rightarrow 0$. Right? Feb 20, 2019 at 1:35
• @user1620696 does my answer look correct? Feb 20, 2019 at 1:43
• @guimption, well, I confess I don't understand what you mean by setting $N =1$. Anyway, your approach is to use the squeeze lemma right? Given sequences $(a_n),(c_n)$ both converging to $L$, if $(b_n)$ is a sequence with $a_n\leq b_n\leq c_n$ then $(b_n)$ converges to $L$ as well. This is equivalent to what I did in the answer, though I recoursed to the convergence definition using $\epsilon$ and $N$. I did that because the OP asked, and because it is good to really understand the idea on how to prove convergence using the definition.
– Gold
Feb 20, 2019 at 1:49
• I see what you are saying. The comparison lemma is essentially a trivial corollary of the squeeze lemma, yes. and I use $N=1$ as per the OP's definition of $N$. Feb 20, 2019 at 2:14

Your professor is not concluding that $|2n|<\varepsilon n^2$, but is instead noting that it is sufficient to have $|2n|<\varepsilon n^2.$

It's certainly not true for all $n,$ but when it is true, you will have $\left|\frac{2n-1}{n^2}\right|<\varepsilon.$ In particular, it is true whenever we have $n>\frac2\varepsilon.$