I came across the following exercise and wanted to check whether my approach was wrong or correct.
(I have read the duplicated here Convergence and limit of a recursive sequence $x_{n+1}=\sqrt{p+x_n}$ but was wondering If I can also sort-of prove it with my method presented here:
Let $$ a_{n} = 1 , a_{n+1}= \sqrt[3]{a_{n}+6} $$ be a recursivley defined sequence that is supposedly bounded from above by 2. Now, when trying to prove it, can I use the fact that $$ \sqrt[3]{a_{n}+6} = (a_{n}+6)^{1/3} $$ and thus $$\sqrt[3]{a_{n+n}+6} = (a_{n}+6)^{1/3^{n}}$$ where the exponent of form 1/3 to the power of n will go to zero, thus the whole sequence converges to 1?
Of course, I will then continue the proof by using the definition of upper bound.
It just feels like I messed up some thought of concept or as though my prof will deduct points for this reasoning. Thanks for your efforts guys.
