Can we find a closed-form formula for $a_{n+1}=\frac{a_n}{a_n+1}$, given $a_1=a$? We have a sequence:
 $$a_{n+1}=\frac{a_n}{a_n+1}$$
$$
a_1 = a
$$
$$
a_2 = \frac{a}{a+1}
$$
$$
a_3 = \text{here things get really messy and for all the following } 
$$
I can't get the formula since the expressions just get more and more complicated. Is there a faster way to do this?
I also have two side questions: What can we say about convergence of this series at the asymptote at $a_n$ = -1 
The other question that I have is when we test the convergence for other values how much should we prove that $a_n$ is increasing/decreasing and how much is enough to just state for instance $a_{n+1} < a_n$ without the proof of it. 
 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.
A: 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.
$$ a_n = \frac{1}{\frac{1}{a}+(n-1)}\,.$$ 
