# Show that $\sum_{n=1}^\infty x_n$ diverges for a sequence defined by $x_{n}=h(x_{n+1})$ [closed]

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.
• Can you show that your $x_{n+1}\geq x_n$ for all $n$? – Stefan Lafon Feb 22 at 5:33
• Sorry, I make a mistake in the questions. The sequence is $x_{1}=a_{1}$ and $x_{n}=h(x_{n+1})$ – Akam2310 Feb 22 at 5:42
• 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. – user21820 12 hours ago
• 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. – Akam2310 23 mins ago

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 $$\sum_{n=1}^\infty x_n = \infty$$.

• $\frac{K}{2+Kx_{n+1}} < \frac{K}{2}$ , 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 – yolomorphism yesterday
• @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 $\frac{1}{1/x_1 + (n-1) K/2} \sim \frac 2 K \frac 1 n$. – Martin R yesterday
• O get it, thank you! – yolomorphism yesterday

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.
• It is $x_{n}=h(x_{n+1})$, not $x_{n+1}=h(x_{n})$. – Martin R 2 days ago