I have to discuss the limit of the following recursive sequence according to the value of $a$. $$\begin{cases}a_1=a;\\ a_{n+1}=a_n\sqrt{\frac{n}{n+1}}\end{cases}$$ I have considered that for $a=0$ the sequence is identically equal to $0$ $\forall n\geq 1$ and so trivially the limit is $0$.
When $a>0$ I have observed that $a_{n+1}-a_n<0$, since if $a>0$ then $a_n>0$ $\forall n\geq 1$. So the sequence is decreasing and the limit coincides with its infimum that is $0$. Similarly, when $a<0$ the sequence is increasing and the limit is its supremum, which is still $0$.
Is my idea right? I have not shown all steps for proving what I have said in the previous lines, since my doubt is in the scheme adopted and not in the specific proofs about increase or decrease of the sequence.
EDIT: I have tried to find the explicit form of $a_n$. I have obtained: $a_n=a(\frac{n}{n+1})^{\frac{n-1}{2}}\to \frac{a}{\sqrt{e}}$...so one of my two ideas is wrong...
Let $$b_n=a_n\sqrt{n}$$ Then $$b_n=b_{n-1}=\dots=b_1$$ Thus $$a_n=\frac a {\sqrt n}$$ And the limit is $0$.