Proof verification of $x_n = \frac{n^2 - 2n}{n+1}$ diverges using $\varepsilon$ definition.

Question

Given $$ x_n = \frac{n^2 - 2n}{n+1} $$ Prove that $\{x_n\}$ diverges.

I've started as follows. Suppose there sequence converges to some value. That means:

$$ \lim_{n \to \infty}\frac{n^2-2n}{n+1} = L $$

Consider $x_n$: $$ \frac{n^2-2n}{n+1} = \frac{n(n-2)}{n+1} < \frac{n^2}{n+1} < \frac{n^2}{n} \le n $$

Now by definition of the limit we want:

$$ |x_n - L| =\left|\frac{n^2 - 2n}{n+1} - L\right| <|n-L| < \varepsilon $$

Suppose $\varepsilon = 1$, then:

$$ |n-L| < 1 $$

But since $L$ is fixed we may by Archimedean principle find $n$ such that $n > L \implies n + 1 > L$ and therefore the limit doesn't exist.

I'm kindly asking to verify my proof or correct me if it's wrong.

Answer 1

You can do it like this: Fix an $L>0.$ $x_n>L\Rightarrow n^2-2n>nL+L\Rightarrow n^2-(L+2)n+(L/2+1)^2>L+(L/2+1)^2\Rightarrow (n-L/2-1)^2>L+(L/2+1)^2 \Rightarrow n>L/2+1+\sqrt{L+(L/2+1)^2} .$

So, for an arbitrary $L$ if we choose $n_0=[L/2+1+\sqrt{L+(L/2+1)^2}]+1$, then $x_n>L$ for all $n\geq n_0$. Hence $x_n\rightarrow \infty.$

Answer 2

You are proving that the sequence doesn't converge but it is not sufficient since the limit could also oscillate in general.

To prove that it diverges, recall that by the definition of divergent limit, for any $M\in \mathbb{R}$ we need to show that eventually

$$\frac{n^2 - 2n}{n+1} \ge M \iff n^2-(2+M)n-M \ge 0$$

Answer 3

The error lies in this step:

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

You know that $\frac{n^2 - 2n}{n+1} - L <n-L$, but does that imply $|\frac{n^2 - 2n}{n+1} - L| <|n-L|$? In other words, does $a<b$ imply $|a|<|b|$?