Calculus
# Limits of Sequences and Series

A sequence adheres to the recursion relation $a_{n}=\sqrt[3]{a_{n-1}a_{n-2}a_{n-3}}$ with initial terms $a_{0}=1,a_{1}=2,a_{2}=1$.

As $n$ tends to infinity, what is the limit of $a_n$ correct to 2 decimal places?

Evaluate

$\sqrt{\dfrac{1}{2}}\times\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}}\times\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2} + \dfrac{1}{2}\sqrt{\dfrac{1}{2}}}}\cdots$

The sequence $1, 2, 3, 2, \ldots$ has the property that every fourth term is the average of the previous three terms. That is, the sequence $\{a_n\}$ is defined as $a_1 = 1, a_2 = 2, a_3 = 3$, and $a_{n+3} = \dfrac{a_{n+2} + a_{n+1} + a_n}{3}$ for any positive integer $n$.

This sequence converges to a rational number $\frac{a}{b}$, where $a$ and $b$ are coprime positive integers. Find $a+b$.