Proof by absurd that a sequence is not bounded above

Question

I'm kind of stuck with a proof.

I have the following sequence $$(u_n) \in \mathbb R ^{\mathbb N}$$ $$u_0 \in \mathbb R ^{+*}$$ $$u_{n+1} = u_n+\frac{n}{u_n}$$ I'm asked to show by absurd that this sequence is not bounded above (without using the fact that its limit is +infinity because I will then have to use that fact to deduce its limit).

Any hint would be greatly appreciated!

Answer 1

Suppose it is bounded above, then as the sequence is increasing it will converge to some $M$, thus for $0<\epsilon<1$ there exists $N \geq M$ such that for all $n \geq N$ $$M-\epsilon < u_n <M.$$ However we then have $$u_{n+1} = u_n + \frac{n}{M} > M - \epsilon + \frac{M}{M} = M +1-\epsilon >M,$$ a contradiction.

Answer 2

Suppose that there is $c>0$ such that $u_n \le c$ for all $n$.Then:

$u_n+\frac{n}{u_n}=u_{n+1} \le c$ for all $n$. Since $u_n>0$ for all $n$ we derive

$u_n^2-cu_n+n \le 0$ for all $n$.

Now define $f_n(x)=x^2-cx+n$. Then we have $f_n(u_n) \le 0$ for all $n$.

But if $n>\frac{c^2}{4}$, then $f_n$ has no zero. Since $f_n(0)=n>0$ it follows that

$f_n(x)>0$ for $n>\frac{c^2}{4}$ and all $x$. Then we get the contradiction $f_n(u_n)>0$ for $n>\frac{c^2}{4}$

Answer 3

Since $u_n \ge 0, \forall n \in \mathbb{N}$, we can use AM-GM to get

$$u_{n+1} = u_n + \frac{n}{u_{n}} \ge 2\sqrt{u_n \cdot \frac{n}{u_{n}}} = 2\sqrt{n}$$

Thus $(u_n)_n$ is clearly unbounded.