# Prove that sequence is divergent

Define the sequence $u_{n}$ as follows : $u_{0}=a>0$, $\forall n \in \mathbb{N}$ , $u_{n+1}=\displaystyle\sqrt{\sum_{k=0}^{n}u_{k}}.$ Prove that the sequence $u_{n}$ diverges.

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Have you tried something? If you work out the first few terms as a function of $a$ you might see something come up. –  Patrick Da Silva Dec 11 '11 at 9:32
If $a=1$, $u_i \ge \sqrt{i}$ does not converge. –  Peteris Dec 11 '11 at 9:50
In light of Martin's answer, you may consider changing the question (or command, rather). –  David Mitra Dec 11 '11 at 9:59
Sorry, edited now. –  user20010 Dec 11 '11 at 10:02

This is essentially Martin's answer again:

A sequence $\{a_n\}$ of positive terms converges if and only if the sequence $\{a_n^2\}$ converges.

We will show that $\{u_n^2\}$ diverges.

Here, the sequence $\{u_n\}$ is increasing for $n\ge1$ (since $\{u_n^2\}$ is increasing) and consists of positive terms. Thus, $$u_{n+1}^2\ =\ u_0+\sum_{k=1}^n\, u_k \ \ge\ u_0+\sum_{k=1}^n \, u_1 \ =\ a +n\cdot \sqrt a \ \ \buildrel{n\rightarrow\infty}\over\longrightarrow\ \ \infty.$$

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Alternatively, you could also use that the root function is a strictly monotonous function and that obviously all $u_{n,n\in\mathbb{N}}>0$ due to the monotony of the root function and the fact that $u_{0}>0$. Then, one has to define an upper bound, because even if the sequence is increasing, it is not guaranteed that there doesn't exist an upper bound. Therefore you can use the following estimate: $u_{n}>u_{0}=a$ hence $u_{n+1}=\sqrt{\sum_{i=0}^{n}u_{i}} >= \sqrt{\sum_{i=0}^{n}u_{0}} = \sqrt{n\cdot a} = C\cdot\sqrt{n}$, where $C=\sqrt{a}>0$. But it is known that $\sqrt{n}$ is a divergent sequence. As the sequence in the exercise is bounded from below by a divergent sequence, it cannot possess a finite upper bound. Hence it must diverge. Q.E.D.

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Hello and thank you for your help, Here is my proof based on your suggestions : Clearly $u_{n}$ is positive for all nonnegative integers $n$. Furthermore, $u_{n+1}=\sqrt{u_{n}+(u_{n}^2})}$ and thus $u_{n+1} \geq \sqrt{(u_{n})^2}=u_{n+1}$, hence $u_{n}$ is increasing. Hence : for all positive integers $n$ $u_{n} = \sqrt{\sum_{k=0}^{n-1}u_{k}}} \geq \sqrt{\sum_{k=0}^{n}u_{0}$ the last being equal to $\sqrt{n} \times \sqrt{u_{0}}$. Since $u_{0}$ is positive it follows that our sequence tends to infinity as n $\to$ infinity, and thus $u_{n} \ \rightarrow \infty$, which means our sequence diverges. –  user20010 Dec 11 '11 at 11:58

It is easy to see that all terms of this sequence are non-negative.

Notice that, for $n\ge 1$, $u_{n+1}=\sqrt{\sum\limits_{k=0}^{n}u_{k}}=\sqrt{u_n+\sum\limits_{k=0}^{n-1}u_{k}}=\sqrt{u_n+u_n^2}$.

This implies $$u_{n+1}^2=u_n+u_n^2.$$ Thus the sequence $(u_n^2)_{n=1}^\infty$ is increasing and, consequently, so is $(u_n)_{n=1}^\infty$.

Moreover, we have $u_{n}\ge\sqrt{a}>0$ for $n\ge 1$, which implies that the sequence $(u_n^2)$ is unbounded. (Since we have $u_{n+1}^2-u_n^2\ge\sqrt a$ and this implies $u_{n+1}^2 \ge u_1^2 + n\sqrt{a}$.)

Consequently, $(u_n)$ is unbounded.

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If $u_{n} \geq \sqrt {a}$, what guarantees us that $(u_{n^2})$ is unbounded ? $a$ is supposed to be a fixed positive real number. However, the beginning of your proof is for me absolutelety correct and can be completed by the what David Heider gives us below. –  user20010 Dec 11 '11 at 10:40
I've tried to give a more detailed explanation of what I had in mind when I was writing the answer. Of course, there are many possibilities how to proceed. –  Martin Sleziak Dec 11 '11 at 10:46