Proving a property that holds for any convergent sequence of real numbers. Suppose a sequence $(a_n)_{n\in\mathbb{N}}$ of real numbers converges to $A > 0.$ I am trying to show that for large enough $n$, $n^{2}+a_nn > (n+ca_n)^2$ where $0<c<1/2$. Manipulating this inequality yields $a_nn(1-2c)>c^2a_n^2$ and for large enough $n$, $a_n$ will be sufficiently close to $A$ and hence nonzero. Dividing both sides by $a_n$ for $n$ thus large enough yields a further reduced form of the desired inequality. How can I formally proceed to show this reduced inequality, or perhaps more directly show the original inequality?
 A: Let $\epsilon=A/2$ and find $N$ be such that for $n>N$, $$0<A/2=A-\epsilon<a_n<A+\epsilon$$
Then if also:
$$n>c^2\frac{3}{2}\frac{A}{1-2c}$$
Then $$n(1-2c)a_n > c^2(A+\epsilon)a_n> c^2a_n^2$$
So, you need $n>M=\max\left(N,c^2\frac{3}{2}\frac{A}{1-2c}\right)$.
Basically, you just need an upper bound on $a_n$ for $n>M$ and that $a_n>0$ for $n>M$. So the choice of $\epsilon$ was arbitrary, and could have been any value in $(0,A]$.
You really don't need $a_n$ converges, you only need:
$$\liminf a_n >0, \limsup a_n<+\infty.$$
A: A direct approach based on your intuition can be made rigorous if you allow asymptotic manipulation as follows.
As $n \to \infty$:
  $n(1-2c) > c^2 a_n$ because $1-2c > 0$ and $a_n \to A$.
  Thus $n > 2c n + c^2 a_n$.
  Also $a_n > 0$ [eventually] because $A > 0$.
  Thus $n a_n > (2c n + c^2 a_n ) a_n = 2c n a_n + c^2 a_n^2$.
A: Rearranging, you wish to show that $\tag1a_nn-2nca_{n}-c^{2}a_{n}^{2}>0$ for large enough $n$. 
Since $a_n\to A>0$ as $n\to \infty $, we may choose $N\in \mathbb N$ such that $n>N\Rightarrow a_n>0$. Then $(1)$ becomes 
$\tag2 n(1-2c)-c^{2}a_n>0$
or
$\tag3n(1-2c)>c^{2}a_n$ 
Now, $1-2c>0$, so LHS is positive and grows without bound as $n\to \infty $ while RHS is bounded by $c^{2}A+1$ for large enough $n$, and now the result follows because we may choose $N_1$ so that $n>N_1\Rightarrow c^{2}a_n<c^{2}A+1$. Then, as soon as $n>\frac{c^{2}A+1}{1-2c}$ we have 
$\tag4n(1-2c)>c^{2}A+1>c^{2}a_n$ 
which will hold for all $n>\max \left \{ N,N_{1},\left \lceil \frac{c^{2}A+1}{1-2c} \right \rceil \right \}$
