Existence of limit of a sequence

Question

Let the following sequence $$4a_{n+1}=(64a_n^3+15)^{1/3}$$the question is

does this sequence converge?

my attempt:

$a_n$ is convergent if and only if $b_n=4a_n$ is so. Therefore $$b_{n+1}=(b_n^3+15)^{1/3}$$ if any limit exist there must be $$l=(l^3+15)^{1/3}$$or $$l^3=l^3+15$$ which has no answer in real numbers but this doesn't provide sufficient condition (it is only necessary condition). then how do i show this in the right manner?

Answer 1

If $(a_n)$ converges, then so does $(c_n)$ where $c_n=64a_n^3$. But $c_{n+1}=c_n+15$, so $(c_n)$ does not converge.

Answer 2

A necessary condition has to be fulfilled, if it is not, then the statement cannot hold. In your particular problem, the limit (if it exists), needs to be an element of the real numbers. You did a proof by contradiction, you assumed the limit would exists which lead to a false statement which then yields to the conclusion, that your assumption was wrong (contradiction).

So assume it exists, let's say $\lim_{n\to\infty} a_n=a$, which gives $$ 4a=(64a^3+15)^{1/3}\iff 64a^3=64a^3+15\iff 0=15 $$ which is the contradiction.

Answer 3

Let $b_n:=64a_n^3$. Then the recurrence is

$$b_{n+1}=b_n+15$$ and there is no doubt about divergence.