Complex sequence converging to $i$

Question

let $a_{n}=\frac{\sqrt{n}-i}{\sqrt{n}+i}, n \in \mathbb N$

and required to prove that $a_{n} \to 1 , n \to \infty$, using only the definition of convergence.

My thinking: Evaluate $|a_{n} -1|$ and eventually use the axiom of archimedes to find an $N$ for any $\epsilon$, such that $\forall n >= N: |a_{n}-1|<\epsilon$. Unfortunately, I seem to arrive at $|\frac{-2i}{\sqrt{n}+i}|$, which does not help me further in regards to axiom of archimedes as it is still complex. Any ideas?

Answer 1

Hint: $| \frac{-2i}{\sqrt{n}+i} | = 2 |\frac{1}{\sqrt{n}+i}|=2 \sqrt{\frac{1}{(\sqrt{n}+i)(\sqrt{n}-i)}} = \frac{2}{\sqrt{n+1}}$

Answer 2

$$ \frac{\sqrt{n}-i}{\sqrt{n}+i}=\frac{\sqrt{n}\left(1-\frac{i}{\sqrt{n}}\right)}{\sqrt{n}\left(1+\frac{i}{\sqrt{n}}\right)} $$ $$ \left|-\frac{i}{\sqrt{n}}\right|=\left|\frac{i}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}\underset{n \rightarrow +\infty}{\rightarrow}0 $$ Hence

$$ a_n=\frac{\sqrt{n}-i}{\sqrt{n}+i} \underset{n \rightarrow +\infty}{\rightarrow}1 $$

Answer 3

Start by breaking the sequence into real and imaginary parts:

$a_n=\displaystyle{\frac{\sqrt{n}-i}{\sqrt{n}+i}=\frac{(\sqrt{n}-i)^2}{n+1}=\frac{n-1}{n+1}+i\frac{2\sqrt{n}}{n+1}}$. I suppose you're able to see why the sequence of the real parts tends to $1$ and why the sequence of the imaginary parts tends to $0$ (if not, for the latter, try dividing both parts of the fraction with $\sqrt{n}$).

Hope this helps!