Prove that a sequence tends to positive infinity if it is increasing and it is not bounded I am asked to prove that $\left(a_{n}\right)_{n=0}^{\infty}$ where $a_{n+1}=\sqrt{a_{n}^{2}+a_{n}}$ tends to positive infinity as n tends to infinity.It is given that $a_{0}>0$
In order to do so,I am asked to prove it's increasing and that it is not bounded.
I have proven that it is increasing by stating $a_{n+1} \geq a_{n}$ and replacing the definition of an element of the sequence.
But I don't know how to do the other two parts.I tried proving it was unbounded through contradiction with the formal definition of a bounded sequence but I got stuck.
 A: As the sequence is increasing, $\lim_n a_n=\infty$ or $\lim_n a_n=M\in\mathbb{R}$. If the sequence is bounded, then $\lim_n a_n=M$, but
$$
M=\lim_n a_n=\lim_n a_{n+1}=\lim_n\sqrt{a_n^2+a_n}=\sqrt{M^2+M},
$$
so $M=0$, but this is a contradiction because the sequence is increasing and $a_0>0$.
A: In the long run, we expect the sequence to grow approximately linearly: when $a_n$ is large, we have $$a_{n+1} = \sqrt{a_n^2 + a_n} \approx \sqrt{a_n^2 + a_n + \frac14} = a_n + \frac12.$$
But we can see almost the same behavior earlier on; for example, try figuring out when $\sqrt{a_n^2 + a_n} \ge a_n + \frac13$, and you will see that this is true whenever $a_n \ge \frac13$.
So the sequence is definitely unbounded if it ever reaches $\frac13$. To get to that point, use the fact that $a_{n+1} \ge \sqrt{a_n}$.
Alternatively, rather than use $\frac13$, you can find the condition under which $\sqrt{a_n^2 + a_n} \ge a_n + \epsilon$, and pick an $\epsilon$ small enough (depending on $a_0$) that this condition holds from the start. This will also prove that the sequence is unbounded: if $a_{n+1} \ge a_n + \epsilon$, then $a_n \ge n \epsilon$, which is unbounded.
A: Let $a_{n}$ be any sequence satisfying $a_{n+1} = \sqrt{a_{n}^2+a_{n}}$. Assume such a sequence is bounded. Since $\{a_{n} \,|\, n \in \mathbb{N}\}$ is a bounded set of real numbers, it has a supremum, $S$. By definition of the supremum,
$$\forall \delta > 0 ,\,\exists n \in \mathbb{N}:\quad a_n>S-\delta.$$
Note that $\sqrt{S^2+S}>S$. Since $x \to \sqrt{x^2+x}$ is continuous at $S$,
$$\forall\varepsilon>0,\,\exists \delta>0:\quad |x-S|<\delta \implies |\sqrt{S^2+S} - \sqrt{x^2+x}| < \varepsilon.$$
Choose $2\varepsilon = \sqrt{S^2+S}-S$. Then choose $\delta$ as above by the continuity statement. Then choose $n$ from the supremum statement. This $n$ has $a_{n+1}>S$, which is a contradiction.
A: I will sketch an inelegant method here. When we take the square-root of a quadratic term, we can suppose that it behaves somewhat like a linear term. With that in mind, suppose that:
$$a_{n+1} = \sqrt{a_n^2+a_n} \ge a_n + c_n$$
where $c_n$ is the maximal value that satisfies this inequality. This becomes a quadratic in $c_n$ which you can solve -- or you might be able to spot the solution by inspection. Then, you can show that $c_n$ is an increasing positive sequence, so in particular, $c_n \ge c_0, \forall n$.
Can you finish the argument from here?
