${\displaystyle (a_n)}$ is a sequence with ${\displaystyle a_n = \frac{1}{\sqrt{n}}}$. Proove that ${ \displaystyle \lim\limits_{n\to\infty}{a_n} = 0 }$, using epsilon-delta method.
First of all, I assume that ${ \displaystyle n \in [1;+\infty) }$.
For all ${n}$, which are larger or equal to ${1}$, ${n^{th}}$ term of sequence is strictly larger than zero and less or equal to one.
${ \forall n \geq 1 : 0 < a_n \leq 1 }$
Assume some ${ \displaystyle \epsilon > 0 }$, such that ${ \displaystyle 1 - \frac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow \left(\frac{1}{1-\epsilon}\right)^2 > n }$
Now let ${ \displaystyle N = 1 + \left\lceil \left(\frac{1}{1-\epsilon}\right)^2 \right\rceil }$, then ${ \displaystyle \left| \frac{1}{\sqrt{n}} - 1 \right| < \epsilon }$ ${ \displaystyle \forall n \geq N }$
Have I understood the task? Is my solution correct?