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Show the sequence isn't bounded: $a_1=1$, $a_{n+1}=a_n+\frac 1 {a_n}$.

Proof by contradiction: Let $M>0$ such that $\forall n: |a_n|< M$.

Let $\epsilon >0 $ and for some $n=N, \epsilon: a_N=M-\epsilon<M $ pluging that in the recursion: $a_{N+1}=M-\epsilon+\frac 1 {M-\epsilon}>M>M-\epsilon$. Contradiction.

I wondered if I could suppose about the boundary that $\forall n: |a_n|\le M$ ? The proof would basically be the same only I could drop the epsilon.

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    $\begingroup$ How do you know that $M-\epsilon + \dfrac{1}{M-\epsilon} > M$? $\endgroup$ – JimmyK4542 Jun 21 '14 at 22:04
  • $\begingroup$ It is enough to take $\epsilon =1/M$ to have such inequality. But you need $M=\sup_{n\in \Bbb{N}}\{a_n\}.$ $\endgroup$ – mfl Jun 21 '14 at 22:09
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Hint: The sequence is obviously increasing. So if it is bounded, then it has a limit $b\ge 1$. Thus $b=\lim_{n\to\infty} a_{n+1}=\cdots$.

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  • $\begingroup$ So $b=b+1/b\implies b=0$ ? $\endgroup$ – GinKin Jun 21 '14 at 22:12
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    $\begingroup$ $b = b + 1/b$ means that $0 = 1/b$. $\endgroup$ – JimmyK4542 Jun 21 '14 at 22:13
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    $\begingroup$ So $b=b+1/b$, which is impossible, contradicting the assumption that the sequence is bounded. $\endgroup$ – André Nicolas Jun 21 '14 at 22:13
  • $\begingroup$ @JimmyK4542 I multiplied both sides by $b^2$. It still contradicts $b\ge 1$. $\endgroup$ – GinKin Jun 21 '14 at 22:14
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Notice $$a_{n+1} = a_n + \frac{1}{a_n} \quad\implies\quad a_{n+1}^2 = a_n^2 + 2 + \frac{1}{a_n^2} \ge a_n^2 + 2$$ This implies for any $n$, $$a_n^2 \ge a_1^2 + 2(n-1) = 2n-1$$ As a result, $a_n$ diverges at least as fast as $\sqrt{2n-1}$ as $n\to\infty$.

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    $\begingroup$ I like this one, it has explicit size information. $\endgroup$ – André Nicolas Jun 21 '14 at 22:17
  • $\begingroup$ Why does it imply $a_n^2 \ge a_1^2 + 2(n-1) = 2n-1$ ? $\endgroup$ – GinKin Jun 22 '14 at 10:16
  • $\begingroup$ @GinKin $$a_{k+1}^2 - a_k^2 \ge 2,\;\forall k \ge 1 \implies a_n^2 = a_1^2 + \sum\limits_{k=1}^{n-1} (a_{k+1}^2 - a_{k}^2) \ge 1 + \sum\limits_{k=1}^{n-1} 2 = 1 + 2(n-1)$$. $\endgroup$ – achille hui Jun 22 '14 at 20:17
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Somewhat of a funny proof.

Suppose for contradiction that $a_n$ is bounded by some positive $M$

Then the series $\displaystyle \sum (a_{n+1}-a_n)=\sum\frac{1}{a_n}$ has bounded partial sums (and positive general term).

Thus, the series $\displaystyle \sum\frac{1}{a_n}$ converges, which implies $a_n\to \infty$

Contradiction.

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Hint: First, show by induction that $a_n > 0$ for all $n$

Now, suppose $a_n$ is bounded, i.e. $a_n \le M$ for some $M > 0$.

Then, $\dfrac{1}{a_n} \ge \dfrac{1}{M}$, and so, $a_{n+1} = a_n + \dfrac{1}{a_n} \ge a_n + \dfrac{1}{M}$.

Now, can you show how this yields a contradiction?

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  • $\begingroup$ I can't find the contradiction, the best I got is $M-\frac 1 M \ge a_n$. Can you please show the contradiction ? $\endgroup$ – GinKin Jun 22 '14 at 10:42
  • $\begingroup$ If $a_{n+1} \ge a_n + \dfrac{1}{M}$, then $a_n \ge 1+\dfrac{n-1}{M}$ for all $n \ge 1$. Now do you see the contradiction? $\endgroup$ – JimmyK4542 Jun 22 '14 at 19:49
  • $\begingroup$ Why does it imply that $a_n \ge 1+\dfrac{n-1}{M}$ ? $\endgroup$ – GinKin Jun 22 '14 at 20:05
  • $\begingroup$ $a_2 \ge a_1 + \dfrac{1}{M} = 1+\dfrac{1}{M}$, $a_3 \ge a_2 + \dfrac{1}{M} \ge 1+\dfrac{1}{M} + \dfrac{1}{M} = 1 + \dfrac{2}{M}$, $a_4 \ge a_3 + \dfrac{1}{M} \ge 1+\dfrac{2}{M} + \dfrac{1}{M} = 1 + \dfrac{3}{M}$. Do you see where this is headed? $\endgroup$ – JimmyK4542 Jun 22 '14 at 20:08
  • $\begingroup$ Right, I see now why it implies that but I still can't see the contradiction. We have now $M\ge 1+\dfrac{n-1}{M}$ and even after some algebraic manipulation, I still don't see it. $\endgroup$ – GinKin Jun 22 '14 at 20:19
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In addition to what was said before, you can prove by induction that $\ln n\leq a_n\leq n$. It holds for $n=1,2,3$ and if it holds for $n$, then $$a_{n+1}=a_n+\frac{1}{a_n}\geq a_n+\frac{1}{n}\geq \ln n + \frac{1}{n}\geq \int_1^n \frac{1}{x} \,dx+\int_n^{n+1} \frac{1}{x} dx=\ln (n+1)$$ and similarly, you show that $a_{n}+\frac{1}{a_n}\leq n+\frac{1}{\ln n}\leq n+1$.

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