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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.

Erik Dz
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4 Answers4

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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 $a_0>0$.

Alan
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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.

Misha Lavrov
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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.

preferred_anon
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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?

legionwhale
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  • Hi legionwhale, many thanks for the answer. I have a doubt. I have understood that this will prove that the sequence is unbounded. But how can I use said result to say it tends to positive infinitive ? And do I need to mention the formal definition of a bounded sequence inorder to conclude your proof? – Erik Dz Oct 16 '21 at 15:34
  • @ErikDz Hi, Erik. Bear in mind that the absolute value of terms tending to infinity is exactly the same as being unbounded. Since all of the terms in this sequence are positive and increasing (with unbounded absolute values), the sequence must tend to infinity. If you would like a more formal argument, suppose that $a_n$ does not tend to infinity. Then, there exists an $M$ such that $a_n \le M, \forall n$ (e.g. $M = \sup{a_n}$). Then, show that this leads to a contradiction. – legionwhale Oct 16 '21 at 15:42