First-year analysis question: an increasing sequence My student showed me this question and I'm genuinely having trouble with it. I will list all the parts so that the pre-requisites are clear.
Q. Let $c \in (0, 1/2)$. We define a sequence $(a_n)$ by:
$$
a_1 = c, \qquad a_{n+1} = a_n(1-a_n).
$$
(i) Show that $a_n < \frac{1}{n+1}$ for all $n \in \mathbb{N}$. (this is fine)
(ii) We define a sequence $(\gamma_n)$ by $\gamma_n = na_n$. Show that $\gamma_n$ is increasing. Deduce that $\gamma_n$ converges to some $L \in (0,1]$. (this is fine)
(iii) Show that for every $\varepsilon > 0$ there is some $n_0$ such that for all $n\geq n_0$
$$
\left|a_n - \frac{L}{n}\right| \leq \frac{\varepsilon}{n}. \text{(this is fine)}
$$
(iv) Show that there is some $n_1$ such that for all $n\geq n_1$,
$$
\frac{\gamma_{n+1}}{\gamma_n} \geq 1 + \frac{1-L}{2n}.
$$
Part (iv) is where the trouble arises. I can get close but not exactly there. I imagine some clever choice of $\varepsilon$ helps. Any small hint would be appreciated.
 A: For $L<1$ we need that $(n+1)a_n = \gamma_n + a_n$ satisfies
$$(n+1)a_n \le \frac{L+1}2$$
See that
$$(n+1)a_n = \gamma_n + a_n \stackrel{\text{(i)}}\le \gamma_n + \frac1{n+1} \tag1$$
Since $\gamma_n$ is increasing, we have that $\gamma_n \le L$.
If $L < 1$ chose $\epsilon = 1-L$ and find $n_1$ such that $\frac1{n+1}\le \frac\epsilon2$ then
$$(1) \le L + \frac\epsilon2 = \frac{L + L+\epsilon}2 =\frac{L+1}2$$
And we have
$$\begin{align*}
\frac{\gamma_{n+1}}{\gamma_n} & = \frac{n+1}n \frac{a_{n+1}}{a_n} = \frac{n+1}n \frac{a_n(1-a_n)}{a_n} \\ & = (1 + \frac1n)(1-a_n) =1 + \frac{1-(n+1)a_n}n \\
& \le 1 + \frac{1-\frac{L+1}2}n = \frac{1-L}{2n}
\end{align*}$$
Now if $L=1$, since $\gamma_n$ is increasing, $\frac{\gamma_{n+1}}{\gamma_n} \ge 1 = 1 + \frac{1-L}{2n}$ naturally.
q.e.d.
A: By setting $b_n=\frac{1}{a_n}$ we get $b_1\geq 2$ and
$$ b_{n+1} = b_{n}+1+\frac{1}{b_n-1}, \tag{1} $$
so $b_n\geq b_1+(n-1) \geq (n+1)$. However, by plugging back this inequality into $(1)$, we get:
$$ b_{n+1} \leq b_n + 1 + \frac{1}{n}\tag{2} $$
hence $b_n \leq b_1 + (n-1) + H_{n-1}$, and:
$$ a_n \in\left[\frac{1}{(n-1)+\frac{1}{a_1}+H_{n-1}},\frac{1}{(n-1)+\frac{1}{a_1}}\right].\tag{3}$$
This is tight enough to prove both (iii) and (iv).
