Show that $\sum_{n=1}^\infty x_n$ diverges for a sequence defined by $x_{n}=h(x_{n+1})$ Let  $h:[-1,1]\to \mathbb{R}$ be a $C^2$ function such that $h(0)=0$, $h^{\prime}(0)=1$ and $h^{\prime}(x)&gt;1$ for all $x&gt;0$. 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.
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
$$
\frac{1}{x_{n+1}} - \frac{1}{x_n} \le \frac{1}{x_{n+1}} - \frac{1}{x_{n+1}(1+ K x_{n+1}/2)} = \frac{K}{2+K x_{n+1}} &lt; \frac K2
$$
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 $\sum_{n=1}^\infty x_n = \infty$.
the sequence is monotonically increasing:
$\frac{\mathrm{h}(\mathrm{x})}{\mathrm{x}}=\frac{\mathrm{h}(\mathrm{x})-\mathrm{h}(\mathrm{0})}{\mathrm{x}-0}=\mathrm{h}^{\prime}(\mathrm{y})&gt;1$
where $\mathrm{y} \in(0, \mathrm{x})$.
So $h(x)&gt;x$ for all $x&gt;0$,
therefore $x_{n+1}=h\left(x_{n}\right)&gt;x_{n}$.
Consequently $x_{n} \nrightarrow 0$ and $\sum_{1}^{\infty} x_{n}=\infty$