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?
 A: 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.
A: Your answer is correct. Here is a proof using a supplementary aspect which also simplifies the approach and might be helpful.
Note the convergence behaviour of a sequence $(a_n)_{n\geq 1}$ does not change when changing finitely many elements of the sequence. Keeping this in mind the first $999$ elements of the sequence do not need special consideration.

Proof: Since the convergence behaviour of the sequence $(a_n)_{n\geq 1}$ does not change when changing finitely many values, it is sufficient to study the convergence behaviour of 
  \begin{align*}
(a_n)_{n\geq1000}=\left(\frac{1}{n^2}\right)_{n\geq 1000}
\end{align*}
We claim the sequence converges to zero:
  \begin{align*}
\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{1}{n^2}=0
\end{align*}
Let $\varepsilon>0$. Since 
  \begin{align*}
\left|a_n-0\right|=\left|\frac{1}{n^2}-0\right|=\left|\frac{1}{n^2}\right|<\varepsilon\qquad\qquad\text{for all }\tag{1} n>\frac{1}{\sqrt{\varepsilon}}
\end{align*}
  we can choose the index $N=N(\varepsilon)$ to be
  \begin{align*}
N=\max\left\{\frac{1}{\sqrt{\varepsilon}},1000\right\}
\end{align*}
  to assure that 
  \begin{align*}
\left|a_n-0\right|<\varepsilon\qquad\qquad\qquad \forall n> N
\end{align*}
proving that $(a_n)_{n\geq 1}$ converges to zero.

Note that when using the maximum in $N=\max\left\{\frac{1}{\sqrt{\varepsilon}},1000\right\}$ we effectively skip the first $1000$ elements of the sequence and guarantee this way the validity of the statement (1) for all $n>N$.
