Mathematical Reflections 2015, Issue 1 - Problem U326

Problem:
Find $$\sum_{n=0}^\infty \dfrac{a_n+2}{a^2_n+a_n+1},$$ where $a_0>1$ and $a_{n+1}=\dfrac{1}{3}(a^3_n+2)$ for all integers $n \geq 0$.

Proposed by Arkady Alt.

Solution:

Since $a_0>1$, then it can be proved by induction that $a_n>1$ for all $n \in \mathbb{N}$.
For all $n \geq 0$ we have $$\begin{array}{lll} \dfrac{a_n+2}{a^2_n+a_n+1}&=&\dfrac{(a_n+2)(a_n-1)}{a^3_n-1}\\ &=& \dfrac{(a_n+2)(a_n-1)}{3(a_{n+1}-1)}\\ &=&\dfrac{a^2_n+a_n+1-3}{3(a_{n+1}-1)}\\ &=&\dfrac{\frac{a^3_n-1}{a_n-1}-3}{3(a_{n+1}-1)} \\  &=& \dfrac{\frac{3(a_{n+1}-1)}{a_n-1}-3}{3(a_{n+1}-1)} \\ &=&\dfrac{a_{n+1}-a_n}{(a_n-1)(a_{n+1}-1)}\\&=&\dfrac{1}{a_n-1}-\dfrac{1}{a_{n+1}-1}  \end{array}$$
Therefore, $$\sum_{n=0}^\infty \dfrac{a_n+2}{a^2_n+a_n+1}=\dfrac{1}{a_0-1}.$$