Convergence of sequences in $\mathbb{C}$

Question

I am currently getting into the field of complex numbers, with the imaginary unit $i^2 = -1$ and stuff. At the moment i am looking onto a few sequences in $\mathbb{C}$, regarding convergence.

I have the feeling that $\mathbb{C}$ still is thin ice for me, so:

Can somebody please just give me a feedback weather the following proof is correct?

I want to prove

$a_n = \frac{1+i}{n+i}$ convergent. I am using the quotient criteria. Also i am using the fact that $i^2 = -1$. $0<\theta<1$

$$|\frac{\frac{1+i}{(n+1)+i}}{\frac{1+i}{n+i}}|=$$

$$|\frac{(1+i)(n+i)}{((n+1)+i)(1+i)}|= |\frac{n+i+in-1}{(n+1)+(n+1)i +i-1}|=$$

$$|\frac{n+in+i-1}{(n+1)+in+2i-1}|<\theta$$ $$because:$$

$$n+in+i-1<n+in+2i$$

$$q.e.d.$$

If that is correct, can somebody please just give me a hint for the sequence:

$$a_n=(e^{i\frac{\pi}{4}})^n$$ Which criterion would you suggest looking on?

Thanks for any help.

P.S: If you are cool you can also tell me how get these || to cover the whole fraction ;) Sorry for this formatting.

Answer 1

I can't quite grasp the argument you present for $\frac{1+i}{n+i}$. It looks like you're applying the ratio test, but what is the $\theta$ that appears amid all of the algebra?

It would be easier just to do $$|a_n|=\left|\frac{1+i}{n+i}\right|=\frac{|1+i|}{|n+i|}<\frac{|1+i|}{n} \to 0 $$

For $(e^{i\pi/4})^n = e^{ni\pi/4}$ you'll need to know Euler's formula for $e^{ix}$. Compare the values for $n=0,8,16,24,\ldots$ with those for $n=4,12,20,\ldots$.

Answer 2

For your second sequence, the best way in my opinion is to show that : $$\lim\left(\mathbb{e}^{ina}\right)_{n \geq0}=L, L \in \mathbb{C} \iff a \equiv 0\pmod{2\pi}.$$ ($\implies$ is sufficient, but the other way is quite easy as $a \equiv 0\pmod{2\pi}$ means the sequence is a constant)