Proving the limit of $\frac{1}{n^2+n}$ = 0 using the $\epsilon$ - N definition

Question

So, I'm stuck on this question, and have been working on it for a few hours, probably because I'm not 100% in understanding the definition. But here's the question:

Using the $\epsilon$ - N definition of the limit prove that,

$\lim_{x \to ∞}{1\over(n^2+n)}$ = $0$ (sorry I don't know how to write fractions in limits).

In other words, given $\epsilon$ > $0$, find explicitly a natural number N which satisfies the statement in the definition of the limit.

So the definition I received in lectures is:

Let $(a_n)_1^∞$ ($1$ should be $n=1$, I couldn't get it to work, sorry) be a sequence of real numbers $(a_n)_1^∞$ = {$a_1, a_2, a_3,...$} .

Definition: We write $\lim_{n \to ∞} a_n = a$, and say the limit of $(a_n)_1^∞$ equals $a$, if for every $\epsilon$ > 0 there exists N ∈ $\mathbb{N}$ such that if $n ≥ N$, then $|a_n - a| < \epsilon$.

So far this is what I've done:

|$\frac{1}{n^2+n}$ + 1| < $\epsilon$

$\therefore$ |$\frac{1-n^2+n}{n^2+n}$ + 1| < $\epsilon$

I did try other ways like disregarding the 'n' in the denominator since $n^2$ is more significant then it, but I don't think I'm supposed to do that. I'm really just stuck on what to do next as I don't really understand the process. Any help would be GREATLY appreciated, thanks! :)

Answer 1

In the OP, it was implicitly assumed that the limit was $-1$. This occurred when writing

$$\left|\frac{1}{n^2+n}-\color{red}{(-1)}\right|<\epsilon$$

The $\color{red}{-1}$ should be $\color{blue}{0}$ instead.

To show that $\lim_{n\to \infty}\frac{1}{n^2+n}=0$ we let $\epsilon>0$ be given. Then, we have

$$\left|\frac{1}{n^2+n}-\color{blue}{0}\right|\le \frac{1}{n}<\epsilon$$

whenever $n>N(\epsilon)=\lfloor \frac{1}{\epsilon}\rfloor +1$. And we are done.

Answer 2

Let us use the definition you stated in your post. (BTW, the attempt you made is wrong).

Let $\epsilon>0$. By using the Archimedean Property, we can find $N\in\Bbb N$ such that $\frac{1}{N}<\epsilon$. Thus, if $n\geq N$ then we get $$\bigg|\frac{1}{n^2+n}-0\bigg|=\frac{1}{n^2+n}<\frac{1}{n}\leq \frac{1}{N}<\epsilon.$$ Hence, $$\lim_{n\to\infty}\frac{1}{n^2+n}=0.$$

Answer 3

Given $\epsilon>0$ you need to find an $N$ such that $$ \left|\frac{1}{n^2+n} - 0\right| < \epsilon$$ for all $n>N.$ (The zero in the absolute value is where the limit goes. Not sure where your $+1$ came from, that would mean the limit was $-1$.)

So, you need to choose $N$ large enough so that $$ \frac{1}{n^2+n} < \epsilon$$ for all $n>N.$ (You can drop the absolute value sign since it's positive.)

We know we can do this cause the left hand side gets smaller and smaller as we take $n$ larger and larger (This is just what the limit being zero means intuitively). At a certain point $n=N$ the left hand side will become smaller than $\epsilon$ and it will be that way for all $n$ larger than that. To find the crossover point we just rearrange the inequality and get $$ n^2 +n > \frac{1}{\epsilon}$$ or $$ n^2+n-\frac{1}{\epsilon}>0.$$ For some value of $\epsilon,$ you can sketch this parabola and find the region of $n$ for which the inequality holds true. The crossover value $N$ will be the next integer after the largest solution to $n^2+n-1/\epsilon =0.$

Alternatively, we can see that if $n^2 > \frac{1}{\epsilon}$ ( or even if $n>\frac{1}{\epsilon}$) then $n^2+n > \frac{1}{\epsilon}.$ So it suffices to solve one of the much easier former inequalities. The $N$ you get that way won't be the crossover value, but it will be larger and that's all we need.

Also note that if $\epsilon$ is chosen to be particularly large (say $\epsilon>1$) Then the inequality will hold for all $n$, so you only need to consider $\epsilon$ smaller than $1.$