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I am wondering what I am missing from my proof. I would like to show that the limit of the sequence $$a_{n+1}=\sqrt{2a_n+3},\,\, a_1=4,$$ goes to $\infty$, as $n \rightarrow \infty$. Is there anything that I should add to make my argument more clear?

Proof: Let $M \in \mathbb{R}$. Choose $n_0 \in \mathbb{N}$ such that $a_{n_0} > {(M^2-3)/2}$. Then, for any $M$, we have $\sqrt{2a_n+3} > M$ for $n \geq n_0$.

QED

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  • $\begingroup$ what conditions do you have on the sequence $(a_n)$? $\endgroup$ Oct 29 '14 at 7:59
  • $\begingroup$ The only other information I was given was $a_1=4$. $\endgroup$
    – 9301293
    Oct 29 '14 at 8:01
  • $\begingroup$ is the sequence defined recursively? $\endgroup$ Oct 29 '14 at 8:01
  • $\begingroup$ The text I am using does not explicitly say, but my assumption is yes (this was partly my confusion as well). $\endgroup$
    – 9301293
    Oct 29 '14 at 8:02
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    $\begingroup$ Computing a few terms of the sequence would have spared you this confusion. $\endgroup$
    – user65203
    Oct 29 '14 at 8:46
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This sequence does not tend to infinity and it is not unbounded.

$$ \lim_{n\to\infty}a_n=3. $$

Proof. First we prove that the sequence is decreasing. Clearly, $0<a_2=\sqrt{2\cdots 4+3}<4=a_1$. Inductively, if $a_k<a_{k-1}$, then $\,0<a_{k+1}=\sqrt{2a_{k}+3}<\sqrt{2a_{k-1}+3}=a_{k}$. Hence, $\{a_n\}_{n\in\mathbb N}$ is decreasing and lower bounded by $0$, and thus it converges, say to $a$. But $$ a_n\to a\quad\Longrightarrow a_{n+1}=\sqrt{2a_n+3}\to \sqrt{2a+3}, $$ and hence $a=\sqrt{2a+3}$ or $a^2-2a-3=0$ or $(a+1)(a-3)=0$, and as $a_n>0$, then $a\ge 0$ and hence $a=3$.

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Your proof can only prove that $\{a_n\}$ is unbounded.

Let $a_n=n\cos \frac{n\pi}{2}$. Although it is unbounded, however it don't tends $\infty$.

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  • $\begingroup$ Perfect. That is what I needed. $\endgroup$
    – 9301293
    Oct 29 '14 at 8:10
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    $\begingroup$ wait, actually, didn't I show that for any real number, I can put my sequence value $a_{n+1}=\sqrt{2a_n+3}, a_1=4$ larger than that? Isn't that the definition of divergence? $\endgroup$
    – 9301293
    Oct 29 '14 at 8:20
  • $\begingroup$ Yes. It is divergent, however it doesn't diverges to $\infty$. $\endgroup$
    – Paul
    Oct 29 '14 at 8:25
  • $\begingroup$ @phatty "Isn't that the definition of divergence?" What you wrote in your comment is the definition of divergent to +oo. The proof you suggest in your question, if valid, would show that the sequence diverges to +oo, not only that it is unbounded (and the proof fails in the specific case you are interested in, but this is another topic). $\endgroup$
    – Did
    Oct 29 '14 at 9:01
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What you are missing in your proof: "Let $M \in \mathbb{R}$. Choose $n_0 \in \mathbb{N}$ such that $a_{n_0} > {(M^2-3)/2}$."

Provided such an $n_0$ exists ! It usually doesn't as the sequence is decreasing and $a_n\le4$.

Assuming that $n_0$ can be found amounts to declaring the sequence divergent, a circular argument.

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