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I want to show that the sequence $$ a_{n+1} = \sqrt{ 1 + a_n^2 } $$ is strictly monotone increasing but not bounded.

That it is strictly increasing is simple I think, just $a_n = \sqrt{a_n^2} < \sqrt{1 + a_n^2}$ and using that $\sqrt{\cdot}$ is strictly increasing.

For unbounded, I used $$ a_{n+1} = \sqrt{1+a_n^2} = \sqrt{1 + (1+a_{n-1})} = \ldots = \sqrt{n + a_1^2}. $$ And it follows because $\sqrt{\cdot}$ is unbounded. Is this right, is there any more formal method to proof this, by showing more directly that there could be no upper bound?

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What about to show that $a_n^2$ is unbounded? –  Sigur Feb 4 '13 at 21:55
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Guess this yields the same manipulations as I did... –  Stefan Feb 4 '13 at 21:58
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To be more formal, use induction and show that $a_n \ge \sqrt{n}$ for all $n$. –  Cocopuffs Feb 4 '13 at 22:06
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You could use the fact that since monotone, if bounded there would be a limit, satisfying the interesting equation $x=\sqrt{1+x^2}$. –  André Nicolas Feb 4 '13 at 22:18
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Well, one does not handle the symbol $\infty$ in that way. If the sequence $(a_n)$ was bounded, it would have a limit $a$. Here $a$ is an ordinary number. Then we would have $a^2=a^2+1$, and therefore $0=1$, which is false. This contradiction shows the sequence cannot be bounded. Of course from $a_{n+1}^2=a+a_n^2$, it is trivial to prove unbounded. I mentioned the (correct) $x^2=1+x^2$ approach as a fun alternative. –  André Nicolas Feb 4 '13 at 23:02

1 Answer 1

Define $b_n:=a_n^2$; then for each $n$, $b_{n+1}=1+b_n$, and by induction, it follows that $b_n=b_0+n$, which proves that $\{b_n\}$ is not bounded. Hence $\{a_n\}$ cannot be bounded.

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