# Proving if sequence is bounded

Given a sequence: $$(a_n)_{n\in\mathbb{N}}:=\frac{n^2}{n+1}$$ I proved the divergence (pretty trivial in this case using $lim_{n\to\infty}n=\infty$).
Now I want to prove, that this sequence is not bounded:
I assume an $K\in\mathbb{R}$ exists, such that $K$ is a bound for $\{|a_n| : n\in\mathbb{N}\}$ (this is the definition for bounded sequences I've got to work with).
Intuitionally I know, that there is always an $n\in\mathbb{N}$ for which $|a_n|>K$. Because this is a contradiction, there can't exist such a $K\in\mathbb{R} \Longrightarrow a_n$ is not bounded.
The problem is, intuition is not enough, how could I prove this? I can't possibly set $n=K$ or somehting like that, because $n\in\mathbb{N}$. Or maybe there is a way to write $K$ using $a_n$? Would appreciate it, if you could help me to understand the "standard" procedure for disapproving bounding property of sequences.

• Are you allowed to use long division? – imranfat Feb 5 '18 at 21:11
• If you already proved $a_n \to \infty$, what's left to prove? – Hans Lundmark Feb 5 '18 at 21:13

write $$\frac{n^2}{n+1}=n-1+\frac{1}{n+1}$$

• All answers were good. I just personally like your solution without inequalities, that's why I accepted it. – user3125470 Feb 5 '18 at 21:40
• ok, i will give you a better answer in the future! – Dr. Sonnhard Graubner Feb 5 '18 at 21:41

Hint:  $\;\;\dfrac{n^2}{n+1} \gt \dfrac{n^2-1}{n+1} = \ldots$

Let $n \in \mathbb{Z^+}$:

$a_n = \dfrac{n^2}{n+1} \ge \dfrac{n^2}{2n} =(1/2)n.$

Let $a$ be real , positive.

Archimedes :

There is a $n_0 \in \mathbb{Z^+}$ such that

$n_0 \gt 2a.$

For $n \gt n_0:$

$a_n \ge (1/2)n \ge (1/2)n_0 \gt a$.

• do we need Archimedes? doesn’t induction suffice? – gimusi Feb 6 '18 at 8:16
• Gimusi.The bound is a real number. There exist an n , natural number greater than a (Archimes' Principle). How would you go about this by induction ? If your bound a is a natural number, then using that every natural number has a successor n+1, characteristic of natural numbers, then I guess no need to dig out Archimedes. Correct me if wrong . :)) – Peter Szilas Feb 6 '18 at 9:21

As an alternayive note that

$$\frac{n^2}{n+1}=\frac{n}{1+\frac1n}\ge\frac{n}2$$