I have seen a multiple choice question:
Q. Let $a_1=5$ and define recursively $a_{n+1}=3^{\frac{1}{4}}\left(a_n\right)^{\frac{3}{4}}, \quad n \geq 1$. Then, which of the following statements is TRUE?
(A) $\left\{a_n\right\}$ is monotone increasing, and $\lim _{n \rightarrow \infty} a_n=3$
(B) $\left\{a_n\right\}$ is monotone decreasing, and $\lim _{n \rightarrow \infty} a_n=3$
(C) $\left\{a_n\right\}$ is non-monotone, and $\lim _{n \rightarrow \infty} a_n=3$
(D) $\left\{a_n\right\}$ is decreasing, and $\lim _{n \rightarrow \infty} a_n=0$
I can see that $\left\{a_n\right\}$ is monotone decreasing, by using induction on the hypothesis $\frac{a_{k+1}}{a_k} \geq 1$ started by $\frac{a_2}{a_1} \geq 1$ but while applying the limit on $a_{n+1}=3^{\frac{1}{4}}\left(a_n\right)^{\frac{3}{4}},$ we get $l^4-3l^3=0$ to conclude $l=0,3$ are potential limits, how to prove $\left\{a_n\right\}$ converges and specify the limit?
Can we prove $\left\{a_n\right\} \geq 3$ for all $n$? then it is easy.
Best Answer
Hint (almost solution). Prove by induction that $a_n\ge 3$. Obviously, $a_1=5>3$. Let $a_k\ge 3$, then $$a_{k+1}=3^{1/4}\cdot a_k^{3/4}\ge 3^{1/4}\cdot 3^{3/4}=3.$$