Sequences (Epsilon-N proof)

Question

Is the sequence $a_n = \sqrt{n}/(n-1234.5)$ convergent or divergent?

Take $\lim (a_n) = \lim \sqrt{n}/(n-1234.5)$. Then by L'H, $\lim 0.5n^{-1/2} = 0$ as $n\to \infty$.

Claim: $a_n \to 0$ as $n \to \infty$.

Let $\epsilon > 0$ be arbitrary. Choose $N > 1234.5$ ?? (My attempt) Then for $n > N$ $$\left|{\sqrt{n}\over n-1234.5} - 0\right| = \left|{\sqrt{n}\over n-1234.5}\right| < \left|{n\over n-1234.5}\right|$$

I am stuck here... I don't know how to manipulate this further to show that this is less than $\epsilon$.

Moreover, how do I choose $N$?

Thanks for the help!!!

Answer 1

$n>2469\implies n-1234.5>n/2>0$ $\implies 0<1/(n-1234.5)<2/n\implies$ $0<\sqrt n /(n-1234.5)<2/\sqrt n.$

Answer 2

Let $t\notin \mathbb{N}$ and $a_{n}=\dfrac{\sqrt{n}}{n-t}$ and let $m=E(t)+1$; let's take $n\geq m\rightarrow a_{n}>0$

$|a_{n}|=\dfrac{\sqrt{n}}{n-t}<\epsilon \rightarrow \sqrt{n}<\epsilon n-\sqrt{n}-\epsilon t>0\rightarrow \sqrt{n}>\dfrac{1+\sqrt{1+4\epsilon^{2}t}}{2\epsilon}$

Well: $\epsilon n-\sqrt{n}-\epsilon t=0\rightarrow \sqrt{n}=\dfrac{1\pm \sqrt{1+4\epsilon^{2}t}}{2\epsilon}$

Then it is taken: $n_{0}=\max\{ m, \left( \dfrac{1+\sqrt{1+4\epsilon^{2}t}}{2\epsilon}\right)^{2}\}$