Can we find a closed-form formula for $a_{n+1}=\frac{a_n}{a_n+1}$, given $a_1=a$?

Hint: If you write the 3rd term for example you start to see the pattern $$a_3 = \frac{\frac{a}{a+1}}{\frac{a}{a+1} + 1} = \frac{a}{2a + 1}$$ and the fourth term is $$a_4 = \frac{\frac{a}{2a+1}}{\frac{a}{2a+1}+1} = \frac{a}{3a + 1}$$ Do you see the pattern? You will need an induction argument to prove this obviously.

If $a=0$ then $a_n=0$ for all $n$.

If $a_n\neq 0$ you can write $b_n = 1/a_n$. Then you have simply $b_{n+1} = 1+b_n$.

It means $b_n = \frac{1}{a} + n$ and so $a_n = \frac{1}{\frac{1}{a} + n}$.

From this expression you see that if $a\neq 0$ you will never have $a_n=0$ and that if $\exists m\in \mathbb{N}$ such that $1/a = -m$ then the sequence is not defined for $n\geq m$.

If $-1/a$ is not an integer, then the sequence will tend to $0$ for $n\to \infty$, no matter the value of $a$.

EDIT: My answer assumed that $a_0=a$. If you want to recover the sequence with $a_1=a$, which is what you wrote in your question, you should shift every index, i.e. $$ a_n = \frac{1}{\frac{1}{a}+(n-1)}\,.$$