##\lim_{n \to \infty} ## for the sequence at ##a_n##

[Helly123]

The problem with this solution is, that we still need an argument why $L$ exists at all. Not that we calculate with something, which doesn't exist. Such an argument can be given by the fact that $a_n < a_{n-1}$, and all are positive, so bounded from below by zero. Therefore a limit has to exist, which is a theorem from calculus: The sequence has to go somewhere and space is running out!

yes i agree

 

[willem2]

Y Such an argument can be given by the fact that $a_n < a_{n-1}$, and all are positive, so bounded from below by zero. Therefore a limit has to exist, which is a theorem from calculus: The sequence has to go somewhere and space is running out!

It's even more useful to consider what happens if a_n = L+x. That will both give you a proof of the existence of the limit and an explicit formula for a_n as well.