Proving convergence of a sequence that is defined in two parts. I have the sequence $a_n =  \begin{cases} \frac {1}{n} ; n<1000 \\ \frac {1}{n^2} ; n\ge1000 \end{cases} $
I have to prove that this sequence converges to 0 using the definition 
$\forall ε>0 : \exists N \in \mathbb N : \forall n>N: |a_n-0| < ε$
So from $|a_n-0| < ε$ I get to $N=\lceil \frac{1}{ε} \rceil$ if $ε> 10^{-6}$ and $N = \lceil \frac{1}{\sqrtε} \rceil$ if $ε\le10^{-6}$
Is this correct?

One little simplification is possible. 
Let us take $N=\lceil \frac{1}{ε} \rceil$.
What happens when we pick $ε> 10^{-3}$? We have $N>1000$.
So, for all $n \geq N$:
$\left |a_{n}-0 \right |=\frac{1}{n^{2}}< \frac{1}{n}< \frac{1}{N}< \epsilon $
What if $ε< 10^{-3}$? We have $N<1000$
So, for all $n \geq N$:
If $N\leq n< 1000$: $\left |a_{n}-0 \right |=\frac{1}{n}$
If $n\geq 1000>N$: $\left |a_{n}-0 \right |=\frac{1}{n^{2}}< \frac{1}{n}< \frac{1}{N}$ again
So you actually don't have to define $N$ in two parts.
