Limiting a positive sequence that goes to zero. Let $(a_n)_{n\in \mathbb{N}}$ be a sequence of strictly positive terms with $\lim_n a_n=0$ such that, for a certain $ c>0$ and for all $n \geq 1$, $|a_{n+1}-a_{n}|\leq c a_n^2$ . Prove that exists a $d>0$ such that $n a_n\geq d $ for all $n\geq 1$.
I have no immediate idea on how to approach this problem.
Any thoughts?
Thanks in advance.
 A: Using $a_n\to 0$, find $N$ such that $a_n<\frac1{2c}$ for all $n\ge N$.
Pick $d>0$ such that $d<\frac1{2c}$ and  $d\le na_n$ for $1\le n\le N$. 
Then $na_n\ge d$ for all $n\in \Bbb N$.
This follows by induction on $n$ which we can start at $n=N$.
Indeed, suppose $na_n\ge d$ and $n\ge N$. 
The function $f(x)=x-cx^2$ is strictly increasing on $[0,a_n]$ because there $f'(x)=1-2cx \ge 1-2ca_n>0$.
Therefore
$$ a_{n+1}\ge a_n-ca_n^2=f(a_n)\ge f(d/n)=\frac dn\cdot\left(1-\frac{cd}{n}\right).$$
Now from
$$ \frac{n+1}{n}\cdot\left(1-\frac{cd}{n}\right)>\frac{n+1}{n}\cdot\left(1-\frac{1}{2n}\right)=\frac{(n+1)(2n-1)}{2n^2}=\frac{2n^2+n-1}{2n^2}\ge1,$$
we obtain
$$ (n+1)a_{n+1}\ge \frac{n+1}{n}\cdot\left(1-\frac{cd}{n}\right)\cdot d>d,$$
as desired.
A: Something might be fishy with your problem statement. @Hagen von Eitzen gave a good response for $n\geq N$ but your problem asks for $n \geq 1$. 
Proposing a counterexample: let:
$a_1=1$
$a_2=1$
$a_n=1/(n-1)$, if $n\geq 3$.
$c=1$
The condition $|a_{n+1}−a_n|≤ca^2_n$ is satisfied for all $n$:
For $n=1$:
$|1−1|=0\leq =c \times 1^2 =1$
For $n=2$:
$|1/2−1|=1/2 \leq =1 \times 1^2 =1$
For $n\geq 3$:
$|1/n−1/(n-1)|=1/(n(n-1)) \leq =1 \times(1/(n-1))^2$
However, there exists no such $d$ satisfying the other the inequality: 
For $n=1$:
$1   \times 1\geq 1$
For $n=2$:
$2    \times 1\geq 2$
For $n\geq 3$:
$n \times 1/(n-1)\geq 1$
See that case 2 requires $d$ being at least $2$ where the limit case for n requires it to be at most $1$.
So the case for 2 messes up that there would exist any $d$ for which the inequality $n a_n \geq d$ for all $n\geq 1$.
