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Let $h:[-1,1]\to \mathbb{R}$ be a $C^2$ function such that $h(0)=0$, $h^{\prime}(0)=1$ and $h^{\prime}(x)>1$ for all $x>0$. Take $a_{1} \in (0,1)$. Define the sequence $x_{n}$ by $x_{1} = a_{1}$ and $x_{n}=h(x_{n+1})$ . Prove $\sum_{n=1}^\infty x_n=\infty$.

I tried to do this question but I'm having problems whit the divergence of this series. I know that the limit of the sequence is zero, and I have an idea of how the sequence works geometrically. An idea of how can I prove the divergence would be appreciated.

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Akam2310 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ Can you show that your $x_{n+1}\geq x_n$ for all $n$? $\endgroup$ – Stefan Lafon Feb 22 at 5:33
  • $\begingroup$ Sorry, I make a mistake in the questions. The sequence is $x_{1}=a_{1} $ and $x_{n}=h(x_{n+1})$ $\endgroup$ – Akam2310 Feb 22 at 5:42
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    $\begingroup$ Are you trying to cheat on a test? Your classmates are asking the same question, but that is no excuse for you to do the same. $\endgroup$ – user21820 12 hours ago
  • $\begingroup$ Hi @user21820, actually this is a question of a test, but there is not problem, we can use books, articles o webs like this for try to answer the question, I was trying so hard whit this question, but I dont see how can I solve. Sorry. $\endgroup$ – Akam2310 23 mins ago
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The sequence $(x_n)$ is strictly decreasing and positive. In order to show that $\sum_{n=1}^\infty x_n$ diverges we need some lower bound for the $x_n$.

Let $K = \max \{ |f''(x)| : -1 \le x \le 1 \}$. From Taylor's theorem we have $$ x_n = h(x_{n+1}) = h(0) + h'(0) x_{n+1} + \frac{h''(c)}{2} x_{n+1}^2 $$ for some $c \in [-1, 1]$, and therefore $$ x_n \le x_{n+1} + \frac K2 x_{n+1}^2 = x_{n+1} (1+\frac K2 x_{n+1})\, . $$

It follows that $$ \frac{1}{x_{n+1}} - \frac{1}{x_n} \le \frac{1}{x_{n+1}} - \frac{1}{x_{n+1}(1+ K x_{n+1}/2)} = \frac{K}{2+K x_{n+1}} < \frac K2 $$ for all $n$. Adding these inequalities gives $$ \frac{1}{x_{n}} - \frac{1}{x_1} \le (n-1) \frac K2 $$ or $$ x_n \ge \frac{1}{1/x_1 + (n-1) K/2} \, . $$ By comparison with the harmonic series it follows that $\sum_{n=1}^\infty x_n = \infty$.

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  • $\begingroup$ $\frac{K}{2+Kx_{n+1}} < \frac{K}{2}$ , Can't we take just less than $K$? And why x_n diverges by comparison with the harmonic series if $\frac{1}{\frac{1}{x_1}+(n-1)K/2}$ isn't the harmonic serie? Can you explain it a little bit to me, please? I mean, I know it is right but I just want to understand this part better, thanks.@Martin $\endgroup$ – yolomorphism yesterday
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    $\begingroup$ @yolomorphism: The point is that $\frac{1}{x_{n+1}} - \frac{1}{x_n} $ is bounded above by some fixed number. It does not matter if you take $K/2$ or $K$ as the upper bound. – For the comparison note that asymptotically $\frac{1}{1/x_1 + (n-1) K/2} \sim \frac 2 K \frac 1 n$. $\endgroup$ – Martin R yesterday
  • $\begingroup$ O get it, thank you! $\endgroup$ – yolomorphism yesterday
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the sequence is monotonically increasing: $\frac{\mathrm{h}(\mathrm{x})}{\mathrm{x}}=\frac{\mathrm{h}(\mathrm{x})-\mathrm{h}(\mathrm{0})}{\mathrm{x}-0}=\mathrm{h}^{\prime}(\mathrm{y})>1$ where $\mathrm{y} \in(0, \mathrm{x})$. So $h(x)>x$ for all $x>0$, therefore $x_{n+1}=h\left(x_{n}\right)>x_{n}$.

Consequently $x_{n} \nrightarrow 0$ and $\sum_{1}^{\infty} x_{n}=\infty$

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exhaustmanifold is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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    $\begingroup$ It is $x_{n}=h(x_{n+1})$, not $x_{n+1}=h(x_{n})$. $\endgroup$ – Martin R 2 days ago

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