Recursive sequence with Complex numbers, missing conclusion.

Question

I am solving the following task:

Let $a_1 = \sqrt{2}\sqrt{3}*i, a_{n+1} =\frac{i* a_n}{n+1}$

What can you say about the convergence of $a_n$? I already found out a lot. What i concluded so far, is:

We can also write $a_n$ like this:

$$a_n = \frac{Re(n) +i*Im(n)}{n!}$$

As for the real and imaginary part we can say that

$Re(n) = \sqrt{2} - \frac{n-1}{2}$ for odd n $Re(n) = \sqrt{3} - \frac{n}{2}$ for even n

and $Im(n) = Re(n-1), Re(n) = Im(n-1)*i$

So i think the sequence is convergent, because n! grows way faster than the numerator. But i am missing the exact conclusion, because i still lack of understanding of how to divide the fraction above, what is the result for each n ? Is it a complex number as well?

If i multiply with the numerators conjugation i get:

$$\frac{Re(n) +i*Im(n)}{n!}\frac{Re(n) -i*Im(n)}{Re(n) -i*Im(n)}= \frac{Re(n)^2 +Im(n)^2}{Re(n)*n! -n! *i*Im(n)}$$

Please help me it is just the conclusion i am missing..!! Maybe i make it way to complicated? I appreciate any hints.

Answer 1

$|a_{n+1}|=|a_n|/(n+1)$ so $|a_n|\to 0$ and $a_n\to 0$ too.