<s>[INST] Given a sequence $ (a_{n})_{n \in \mathbb{N}} $, find $ \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $. I would appreciate it if somebody could help me with the following problem:

If the sequence $ (a_{n})_{n \in \mathbb{N}} $ satisfies
  $$
a_{1} = 1
\quad \text{and} \quad
\forall n \in \mathbb{N}, n \ge 2: \quad
a_{n} = \frac{(a_{1} + 2) (a_{2} + 2) \cdots (a_{n-1} + 2)}{2^{n}},
$$
  find $ \displaystyle \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $.

 [/INST] For $n \ge 2$, we have: $a_{n+1} = \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)(a_n+2)}{2^{n+1}}$ $= \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)}{2^n}\cdot \dfrac{a_n+2}{2} = \dfrac{a_n(a_n+2)}{2} = \dfrac{1}{2}a_n^2+a_n$. 
Thus, $\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}} = \dfrac{a_{n+1}-a_n}{a_na_{n+1}} = \dfrac{\tfrac{1}{2}a_n^2}{a_n \cdot \tfrac{1}{2}a_n(a_n+2)} = \dfrac{1}{a_n+2}$. 
Therefore, $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{a_n+2} = \dfrac{1}{a_1+2} + \sum_{n=2}^{\infty}\left[\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}\right] = \dfrac{1}{a_1+2} + \dfrac{1}{a_2} = \dfrac{1}{1+2} + \dfrac{1}{\tfrac{3}{4}} = \boxed{\dfrac{5}{3}}$, 
where we have used the fact that the sum is telescoping and $a_n \to \infty$ as $n \to \infty$.

EDIT: We can show that $a_n \to \infty$ without assuming that the series converges. 
Note that $\dfrac{a_{n+1}+1}{2} = \dfrac{1}{4}a_n^2+\dfrac{1}{2}a_n+\dfrac{1}{2} = \left(\dfrac{a_n+1}{2}\right)^2+\dfrac{1}{4} \ge \left(\dfrac{a_n+1}{2}\right)^2$ for all $n \ge 2$. 
Since $a_3 = \dfrac{33}{32}$, by induction we have $\dfrac{a_n+1}{2} \ge \left(\dfrac{a_3+1}{2}\right)^{2^{n-3}} = \left(\dfrac{65}{64}\right)^{2^{n-3}}$ for all $n \ge 3$. 
Clearly, $\displaystyle\lim_{n\to\infty}\left(\dfrac{65}{64}\right)^{2^{n-3}} = +\infty$. Therefore, $\displaystyle\lim_{n\to\infty}a_n = +\infty$ as desired. 
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